Power Supply Design Seminar
Topic 4 Presentation:
Practical Magnetic Design:
Inductors and Coupled Inductors
Reproduced from
2012 Texas Instruments Power Supply Design Seminar
SEM2000, Topic 4
TI Literature Number: SLUP305
© 2012, 2013 Texas Instruments Incorporated
Power Seminar topics and online power training
modules are available at:
ti.com/psds
Topic 4
Practical Magnetic Design:
Inductors and Coupled Inductors
Louis Diana
SLUP305
Agenda
•
Theory of operation and design equations
– Design flow diagram discussion
– Inductance calculations
– Ampere’s law for magnetizing force
– Faraday’s law for flux density
– BH curves and magnetic saturation
– Core loss
– AC and DC wire losses
– Effective permeability and inductance rolloff
– Coupled-inductor winding layers and considerations
•
Example – flyback-coupled inductor design using EFD core
Texas Instruments – 2012 Power Supply Design Seminar
4-2
SLUP305
Design Flow
Specifications required:
1. Inductance
2. Turns
3. Peak current
4. RMS current
5. Output power
6. Frequency
Specification
Pick a Core Based
on Output Power
Inductance
Calculation
Yes
Wire-Loss Calculations
Yes
Change
Core Size
No Flux Density and CoreLoss Calculations
Yes
No
Bobbin-Fit
Calculations
No
Change
Wire Size
or Filar
Yes
Build and Test
Magnetic
Texas Instruments – 2012 Power Supply Design Seminar
4-3
SLUP305
Magnetic Parameters
and Conversion Factors
SI
CGS
CGS to SI
Flux Density
B
Tesla
Gauss
10-4
Field Intensity
H
A-T/m
Oersted
1,000/4π
Permeability
µ
4 π x10-7
1
4 π x10-7
Area
Ae
m2
cm2
10-4
Length
lg, le
m
cm
10-2
Texas Instruments – 2012 Power Supply Design Seminar
4-4
SLUP305
Inductance
The inductance, L, of a wound core can be calculated from the core
geometry using the following equation:
0.4×π×N 2 ×Ae×10−8
L=
le
lg +
µ
µ = Core permeablity
N = Number of turns
Ae = Core cross - section (mm2 )
le = Core magnetic path length (mm)
lg = Gap (mm)
Manufacturers list inductance for a given core and gap as inductance per turn squared,
referred to as Al .
L = Al×N 2
Texas Instruments – 2012 Power Supply Design Seminar
For Al in nH/per turn 2
4-5
SLUP305
Core Geometry
Mean magnetic path length (le) and effective area (Ae) are given in most data
sheets, but le can also be calculated using the following equation:
OD
ID
le =
π(OD-ID)
OD
)
ln(
ID
Ae
le
Texas Instruments – 2012 Power Supply Design Seminar
4-6
SLUP305
Ampere’s Law
Ampere’s law states that the total magnetic force along a closed path is
proportional to the ampere-turns in the winding that the path passes through.
SI: F =
∫ H×dl = N×I ≈ H×le
H≈
CGS: F =
N×I
ampere-turns/meter(A-T/m)
le
∫ H×dl = 0.4×π×N×I ≈ H×le
0.4×π×N×I
H≈
(Oersteds)
le
H = Magnetizing force
N = Number of turns
le = Core magnetic path length (m for SI, cm for CGS)
I = Peak magnetizing current (amperes)
Texas Instruments – 2012 Power Supply Design Seminar
4-7
SLUP305
Faraday’s Law
The total magnetic flux Φ passing through a surface of area
Ae is related to the flux density β.
ΔΦ =
SI:
1
E×dt
∫
N
E×dt
dΦ
dβ
∫
E = N×
≈ N×Ae× ⇒ β =
dt
dt
N×Ae
β = Flux density (Tesla)
CGS:
10 8
ΔΦ =
E×dt
∫
N
E =
N dΦ N×Ae×dβ
×
≈
⇒β =
8
8
dt
10
10 ×dt
β = Flux density (Gauss)
E = Voltage across coil (V)
8
×10
E×dt
∫
N×Ae
Ae = Effective core area (m 2 )
Texas Instruments – 2012 Power Supply Design Seminar
4-8
SLUP305
BH Curves
β
βmax
βmax
βac
Ηmin
β = Flux density (Tesla)
Ηmax
Np = Number of primary turns
2
Ae = Effective core area (m )
Vin = Voltage (V)
Bac is the AC flux density and
Bmax is the peak flux density.
Vin×ton
βac =
Ae×Np
Tesla
Lp×Ip
βmax =
Ae×Np
Tesla
ton = On time (s)
Lp = Primary inductance (H)
Ip = Peak primary current (A)
Texas Instruments – 2012 Power Supply Design Seminar
4-9
SLUP305
Core Loss
For a Ferroxcube core:
βacpeak =
βac
2
Convert Ve from mm 3 to m 3
Ve(m 3 ) =
Ve
1,000 3
Pcoreloss = Pv(KW m 3 )×Ve(m 3 )
Specific power loss as a function of one-half of
peak-to-peak flux density with frequency as
a parameter.
Courtesy of Ferroxcube
Texas Instruments – 2012 Power Supply Design Seminar
4-10
SLUP305
Permeability and Inductance Rolloff
As current is increased in a magnetic and the flux density gets closer
to core saturation, the permeability of the core starts to roll off.
CBW477
10 4
handbook, halfpage
0.4×π×N 2 ×Ae×10-8
L=
le
lg+
µ
3F3
µ
µ rev
10 3
10 2
10
1
10
10 2
3
H (A/m) 10
Permeability as a function of magnetizing
force (H) (courtesy of Ferroxcube).
Courtesy of Ferroxcube
Texas Instruments – 2012 Power Supply Design Seminar
4-11
SLUP305
AC–DC Wire Loss
Determine wire area and AC losses:
Open
J
Wire area required = Irms / J
J = 400 A/cm 2
Actual
δ(skin depth cm) =
Equivalent
ρ
7.6
≈
at 100°C
π×µ×f
f
Number of wires required =
Wire area required
Area of wire chosen
Ratio of AC to DC losses
Total winding resistance = AC–DC wire resistance of selected wire×Number of turns×
Texas Instruments – 2012 Power Supply Design Seminar
Length per turn
Number of wires used
4-12
SLUP305
AC–DC Wire Loss
Using look-up tables from the “Magnetics Design Handbook (MAG100A)” by
Llyod H. Dixon:
Determine the wire area in cm 2 and the radius in cm
{
Annular inner ring (cm 2 ) = π× radius of wire 2 −[wire radius(cm)−skin depth(cm)]2
}
Area of wire in (cm 2 )
Ratio Rac to Rdc (Rskin) =
Annular inner ring (cm 2 )
AC–DC wire resistance = DC resistance of chosen wire×Ratio Rac to Rdc (Rskin)
Texas Instruments – 2012 Power Supply Design Seminar
4-13
SLUP305
Proximity Effects
AC current in a conductor induces eddy currents in adjacent
conductors by a process called the proximity effect.
P1
P2
S S S
3 2 1
F=Nl
w
Pa
Sa Sb
Pb
F=Nl
w
Power loss Pm in layer m is:
' wire_dia
*
2
× Rdc ,
Pm = I 2 "#( m −1) + m 2 $% × )
(
+
δ
Texas Instruments – 2012 Power Supply Design Seminar
4-14
SLUP305
Winding Factor and
Bobbin Fit Calculations
0
14.8 í 0.2
Winding data and area product for EFD20/10/7 coil
former with 10-solder pads
Number of Sec,ons 1 Winding Area (mm2) Minimum Winding Width (mm) Average Length of Turn (mm) 13.5 34.1 27.7 Area Product Ae x Aw (mm4) Type Number 859 CPHS-­‐EFD20-­‐1S-­‐10P 0
í 0.15
10.3
9.2 + 0.15
0
9.5
max.
3.9
+0.1
0
5 í 0.1
0
Courtesy of Ferroxcube
Winding factor for a bobbin should be in the .3 to .7
range depending on isolation requirements
Turns per layer =
Build up =
1
0.3
7.5 ± 0.05
15 ± 0.05
21.7 max.
Bobbin width (mm)
-2
Diameter of wire in (mm)
0
14.8 í0.2
Winding area in (mm)
Winding width in (mm)
Number of avaliable layers =
Buildup
Diameter of chosen wire
Total bobbin turns = Turns per layer×Number of layers
Winding fill factor =
(13.5 min.)
Turns needed
Total bobbin turns avaliable
0.3
1.8
21
23.7 max.
2
Dimmensions in mm
Texas Instruments – 2012 Power Supply Design Seminar
4-15
SLUP305
Transformer Layers
Transformer layout is very important because it affects
primary to secondary coupling and leakage inductance.
Fourth layer: 32awg single
filar half of primary.
Third layer: 32awg trifilar.
Second layer: 32awg trifilar.
First layer: 32awg single
filar half of primary.
Texas Instruments – 2012 Power Supply Design Seminar
4-16
SLUP305
Design Flow Summary
•
Complete specification
•
Pick a core (see section 4, “Power Transformer Design” by Lloyd H.
Dixon)
•
Calculate inductance and turns-based Al
•
Calculate copper loss
•
Calculate flux density and core loss
•
Calculate temperature rise (see “Constructing Your Power Supply –
Layout Considerations” by Robert Kollman)
•
Iterate as required to get the turns to layer nicely on the core for
good coupling and low leakage inductance
Texas Instruments – 2012 Power Supply Design Seminar
4-17
SLUP305
Flyback-Coupled Inductor Example
Specifications required:
1. Inductance
2. Turns
3. Peak current
4. RMS current
5. Output power
6. Frequency
Specification
Pick a Core Based
on Output Power
Inductance
Calculation
Yes
Wire-Loss Calculations
Change
Core Size
Yes
No Flux Density and CoreLoss Calculations
Yes
No
Bobbin-Fit
Calculations
No
Change
Wire Size
or Filar
Yes
Build and Test
Magnetic
Texas Instruments – 2012 Power Supply Design Seminar
4-18
SLUP305
Flyback-Coupled Inductor Example
Specifications required:
General
Topology
Main Output Power
Maximum Switching Frequency at Full Load
Input
Minimum Input Voltage
Maximum Input Voltage
Primary Peak Current
Primary RMS Current
Outputs
Secondary Output Voltage
Secondary Peak Current
Secondary RMS Current
Bias Voltage
Bias Current
Inductance and Turns Ratio
Primary Inductance
Leakage Inductance
Primary-to-Secondary Turns Ratio
Texas Instruments – 2012 Power Supply Design Seminar
Quasi-Resonant Flyback
10 W
140 kHz
85 VAC, 76 VDC
265 VAC, 375 VDC
1.155 A
0.425 A
5V
13.861 A
5.382 A
16 V
50 mA
190.918 µH
3.818 µH
12
4-19
SLUP305
Flyback-Coupled Inductor Example
•
Core selection is based on
− Core size
− Core material
− Core gap
See section 5 in “Magnetics Design Handbook”
by Lloyd H. Dixon (MAG100A)
•
An EFD20 core with 3F3 material was chosen
− Al was set to 82 nH
− Key core and bobbin data
Winding Area Min. Winding
(mm2)
Width
(mm)
27.7
13.5
Length per
Turn
(mm)
Area
Product
(mm4)
Buildup
(mm)
Effective
Volume
(mm3)
34.1
859
2.052
1460
Texas Instruments – 2012 Power Supply Design Seminar
4-20
SLUP305
Flyback-Coupled Inductor Example
• Calculate primary turns, Np, based on required inductance
and Al value
⎛ Lp ⎞⎟1/2
Np = ⎜⎜⎜ ⎟⎟
⎝ Al ⎠
Lp = 190.918 µH
Al = 82 nH
Np = 48
• Calculate secondary turns-based turns ratio
Ns =
Np
Turns ratio
Turns ratio = 12
Ns = 4
Texas Instruments – 2012 Power Supply Design Seminar
4-21
SLUP305
Flyback-Coupled Inductor Example
• Determine required primary wire area based on current
density, J
Primary wire area =
Ipri_max
J
J = 400 A/cm 2
Ipri_max = 0.425 A
Primary wire area = 1.0625 × 10-3 cm 2
• Determine skin depth at 100°C
δ(skin depth cm) ≈
7.6
f
f = 140 kHz
Skin depth = 0.0203 cm
Texas Instruments – 2012 Power Supply Design Seminar
4-22
SLUP305
Flyback-Coupled Inductor Example
• Calculate annular ring area
( )
{
Annular inner ring cm 2 = π× radius of wire 2 – [wire radius (cm) – skin depth (cm)]
2
}
– If skin depth is greater than wire radius annular ring area = wire area
• Calculate AC to DC resistance, Rac/Rdc ratio
Ratio Rac to Rdc (Rskin) =
Area of wire in (cm 2 )
Annular inner ring (cm 2 )
AWG
Cu Radius
(cm)
Cu Area
(cm2)
Annular Ring
Area (cm2)
Rac/Rdc Rskin
24
0.0255
0.002047
0.001958
1.0453
26
0.02
0.001287
0.001287
1.0000
28
0.016
0.00081
0.000810
1.0000
30
0.0125
0.000509
0.000509
1.0000
Texas Instruments – 2012 Power Supply Design Seminar
4-23
SLUP305
Flyback-Coupled Inductor Example
•
Calculate number of wires required based on wire area,
current density and Rac/Rdc ratio
Numbers of wires required =
Wire area required
Area of wire chosen
Ratio of AC to DC losses
AWG
Wire Area
Required
(cm2)
Cu Area
(cm2)
Rac/Rdc
Rskin
Number of
Wires Required
24
0.0010625
0.002047
1.0453
1
26
0.0010625
0.001287
1.0000
1
28
0.0010625
0.00081
1.0000
2
30
0.0010625
0.000509
1.0000
3
32
0.0010625
0.00032
1.0000
4
Texas Instruments – 2012 Power Supply Design Seminar
4-24
SLUP305
Flyback-Coupled Inductor Example
•
Look at how the windings might fit in the bobbin window
– Initially estimate that half the area is used by the primary
Turns per layer =
Bobbin width (mm)
Diameter of wire (mm)
Number of available layers =
Buildup
Diameter of chosen wire
Buildup =
−2
Winding area in (mm)
Winding width in (mm)
AWG
Insulated
Diameter
(cm)
Min.
Winding
Width (mm)
Turns per
Layer
Available
Layers
Number of
Wires
Required
24
26
28
30
32
0.057
0.046
0.037
0.03
0.024
13.5
13.5
13.5
13.5
13.5
21
27
34
43
54
3
4
5
6
8
1
1
2
3
4
Texas Instruments – 2012 Power Supply Design Seminar
4-25
SLUP305
Flyback-Coupled Inductor Example
Estimate primary winding copper
losses
•
One wire of wire gauge 26 was
chosen
Effective core parameters (courtesy of
Ferroxcube).
Length_per_turn_cm =
34.1 mm
10
Rcu_pri_26awg = Rcu26_100°C×Np×
Length_per_turn_cm = 3.41 cm
Length per turn
Number of primary wires
Rcu_pri_26awg = 0.293 Ω
Pcu_pri_loss_26awg = Ipri_rms 2 ×Rcu_pri_26awg
Pcu_pri_loss_26awg = 0.053 W
Texas Instruments – 2012 Power Supply Design Seminar
4-26
SLUP305
Flyback-Coupled Inductor Example
Ns = 4
Determine secondary wire AWG and losses:
Secondary wire area =
Isecondary
J
Number_of_wires_sec_28awg =
= 14×10-3cm 2
where J = 400 A/cm 2
Area_req_sec_cm 2
Area_28awg_cm 2
Rskin_28awg_100°C
Number_of_wires_sec_28awg = 16.667
Number_of_wires_sec_28awg_used = 5
Rcu_sec_28awg = Rcu28_100°C×Ns×
Length per turn
Number of secondary wires
Rcu_sec_28awg = 7.761×10 -3
Pcu_sec_loss_28awg = Isec_rms 2 ×Rcu_sec_28awg
Pcu_sec_loss_28awg = 0.226 W
Texas Instruments – 2012 Power Supply Design Seminar
4-27
SLUP305
Flyback-Coupled Inductor Example
Determine bias wire AWG and losses:
Number of bias wires = 1
Nbs = 13
Rcu_bias_32awg = Rcu32_100°C×Nsb×
Length per turn
Number of bias wires
Rcu_bias_32awg = 225×10-3
Pcu_bias_loss_32awg = Ibias_rms 2 ×Rcu_bias_32awg
Pcu_bias_loss_32awg = 4.197×10-4 W
Texas Instruments – 2012 Power Supply Design Seminar
4-28
SLUP305
Flyback-Coupled Inductor Example
Determine core flux density:
Ae = 31 mm 2
Np = 48 Ipri_p = 1.155 A
tonmax = 2.9 µs
MBW015
500
-25ºC
-100ºC
B
(mT)
Vinmin×tonmax
Bac =
Ae×Np
400
Bac = 148.17 mT
200
3F3
300
100
Bmax =
Lp×Ipri_p
Ae×Np
Bmax = 148.19 mT
Texas Instruments – 2012 Power Supply Design Seminar
0
-25
0
25
50
150
H (A/m)
250
Courtesy of Ferroxcube
4-29
SLUP305
Flyback-Coupled Inductor Example
Determine core loss:
10
MBW048
4
3F3
T=100 oC
10
3
10
2
25
kHz
Bac
Bunipolar =
2
700
kH
z
400
kH
z
200
kHz
100
kHz
Pv
(kW/m 3)
Bunipolar_mt = 74.1 mT
From graph Pcore = 60 kW/m 3
Ve = 1460 mm 3
Ve_per_m 3 =
Pcore_loss = Pcore×Ve_per_m 3
10
1
Ve
1,000 3
Pcore_loss = 0.088 W
Texas Instruments – 2012 Power Supply Design Seminar
10
10
2
B (mT)
10
3
Specific power loss as a function of
Peak flux density with frequency
as a parameter.
Courtesy of Ferroxcube
4-30
SLUP305
Flyback-Coupled Inductor Example
Magnetic power dissipation:
Pcu_loss_tot = Pcu_pri_loss + Pcu_sec_loss + Pcu_bias_loss
Ptot_mag = Pcu_loss_tot + Pcore_loss
Ptot_mag = 0.367 W
Determine bobbin fit factor:
AWG
Copper Diameter in cm with
Insulation
Copper Area in cm2 with
Insulation
26
0.046
0.001671
28
0.037
0.001083
30
0.03
0.000704
32
0.024
0.000459
bobbin_width_mm = 13.5 mm
bobbin_wdth_cm = bobbin_width_mm×.1
bobbin_width_cm = 1.35 cm
Texas Instruments – 2012 Power Supply Design Seminar
4-31
SLUP305
Flyback-Coupled Inductor Example
Determine bobbin fit factor continued:
Turns_per_layer26 =
bobbin_width_cm
−2
dia_26awg_Inso_cm
Turns_per_layer28 ≈ 34
Turns_per_layer26 ≈ 27
Turns_per_layer30 ≈ 43
Winding_area_mm = 27.7 mm 2
Winding_area_cm 2 = .277
winding_area_cm 2
Build_up_cm =
bobbin_width_cm
Build_up = .205 cm
Layers =
Build_up_cm
dia_26awg_Inso_cm
Layers ≈ 4
Total_bobbin_turns_26awg = Turns_per_layer×Layers
Total_bobbin_turns_26awg =108
Turns_needed = Np×number_of_wires_pri+ Ns×number_of_wires_sec +
Nsb×number_of_wires_bias Turns_needed = 81
Winding_factor =
Turns_needed
Total_bobbin_turns_26awg
Texas Instruments – 2012 Power Supply Design Seminar
Winding_factor = .75
4-32
SLUP305
Flyback-Coupled Inductor Example
Coupled inductor layout:
First layer: 24 turns of 26awg single filar half of primary
Second layer: 20 turns of 28awg five filar secondary plus 13 turns of 30awg bias winding
Third layer: 24 turns of 26awg single filar half of primary
Black lines are tape.
Texas Instruments – 2012 Power Supply Design Seminar
4-33
SLUP305
Summary
•
Discussed theory of operation and design flow
•
Determined inductance, flux density, core loss, AC and DC wire
losses
•
Discussed BH curves and magnetic saturation, effective
permeability and inductance rolloff
•
Discussed transformer winding layers and how to interleave the
winding for good coupling and low leakage
•
Went through an example of a flyback-coupled inductor design
using EFD core
Texas Instruments – 2012 Power Supply Design Seminar
4-34
SLUP305
TI Worldwide Technical Support
Internet
Asia
TI Semiconductor Product Information Center Home Page
support.ti.com
TI E2E™ Community Home Page
e2e.ti.com
Product Information Centers
Americas
Brazil
Mexico
Phone
+1(512) 434-1560
Phone
0800-891-2616
Phone
0800-670-7544
Fax
Internet/Email
+1(972) 927-6377
support.ti.com/sc/pic/americas.htm
Europe, Middle East, and Africa
Phone
European Free Call
00800-ASK-TEXAS
(00800 275 83927)
International
+49 (0) 8161 80 2121
Russian Support
+7 (4) 95 98 10 701
Note: The European Free Call (Toll Free) number is not active in all countries.
If you have technical difficulty calling the free call number, please use the
international number above.
Fax
+(49) (0) 8161 80 2045
Internet
www.ti.com/asktexas
Direct Email
asktexas@ti.com
Japan
Phone
Fax
Internet/Email
Domestic
International
Domestic
International
Domestic
0120-92-3326
+81-3-3344-5317
0120-81-0036
support.ti.com/sc/pic/japan.htm
www.tij.co.jp/pic
The platform bar and E2E are trademarks of Texas Instruments.
All other trademarks are the property of their respective owners.
Phone
International
+91-80-41381665
Domestic
Toll-Free Number
Note: Toll-free numbers do not support
mobile and IP phones.
Australia
1-800-999-084
China
800-820-8682
Hong Kong
800-96-5941
India
1-800-425-7888
Indonesia
001-803-8861-1006
Korea
080-551-2804
Malaysia
1-800-80-3973
New Zealand
0800-446-934
Philippines
1-800-765-7404
Singapore
800-886-1028
Taiwan
0800-006800
Thailand
001-800-886-0010
Fax
+8621-23073686
Email
tiasia@ti.com or ti-china@ti.com
Internet
support.ti.com/sc/pic/asia.htm
Important Notice: The products and services of Texas Instruments
Incorporated and its subsidiaries described herein are sold subject to TI’s
standard terms and conditions of sale. Customers are advised to obtain
the most current and complete information about TI products and services
before placing orders. TI assumes no liability for applications assistance,
customer’s applications or product designs, software performance, or
infringement of patents. The publication of information regarding any
other company’s products or services does not constitute TI’s approval,
warranty or endorsement thereof.
B090712
SLUP305
IMPORTANT NOTICE
Texas Instruments Incorporated and its subsidiaries (TI) reserve the right to make corrections, enhancements, improvements and other
changes to its semiconductor products and services per JESD46, latest issue, and to discontinue any product or service per JESD48, latest
issue. Buyers should obtain the latest relevant information before placing orders and should verify that such information is current and
complete. All semiconductor products (also referred to herein as “components”) are sold subject to TI’s terms and conditions of sale
supplied at the time of order acknowledgment.
TI warrants performance of its components to the specifications applicable at the time of sale, in accordance with the warranty in TI’s terms
and conditions of sale of semiconductor products. Testing and other quality control techniques are used to the extent TI deems necessary
to support this warranty. Except where mandated by applicable law, testing of all parameters of each component is not necessarily
performed.
TI assumes no liability for applications assistance or the design of Buyers’ products. Buyers are responsible for their products and
applications using TI components. To minimize the risks associated with Buyers’ products and applications, Buyers should provide
adequate design and operating safeguards.
TI does not warrant or represent that any license, either express or implied, is granted under any patent right, copyright, mask work right, or
other intellectual property right relating to any combination, machine, or process in which TI components or services are used. Information
published by TI regarding third-party products or services does not constitute a license to use such products or services or a warranty or
endorsement thereof. Use of such information may require a license from a third party under the patents or other intellectual property of the
third party, or a license from TI under the patents or other intellectual property of TI.
Reproduction of significant portions of TI information in TI data books or data sheets is permissible only if reproduction is without alteration
and is accompanied by all associated warranties, conditions, limitations, and notices. TI is not responsible or liable for such altered
documentation. Information of third parties may be subject to additional restrictions.
Resale of TI components or services with statements different from or beyond the parameters stated by TI for that component or service
voids all express and any implied warranties for the associated TI component or service and is an unfair and deceptive business practice.
TI is not responsible or liable for any such statements.
Buyer acknowledges and agrees that it is solely responsible for compliance with all legal, regulatory and safety-related requirements
concerning its products, and any use of TI components in its applications, notwithstanding any applications-related information or support
that may be provided by TI. Buyer represents and agrees that it has all the necessary expertise to create and implement safeguards which
anticipate dangerous consequences of failures, monitor failures and their consequences, lessen the likelihood of failures that might cause
harm and take appropriate remedial actions. Buyer will fully indemnify TI and its representatives against any damages arising out of the use
of any TI components in safety-critical applications.
In some cases, TI components may be promoted specifically to facilitate safety-related applications. With such components, TI’s goal is to
help enable customers to design and create their own end-product solutions that meet applicable functional safety standards and
requirements. Nonetheless, such components are subject to these terms.
No TI components are authorized for use in FDA Class III (or similar life-critical medical equipment) unless authorized officers of the parties
have executed a special agreement specifically governing such use.
Only those TI components which TI has specifically designated as military grade or “enhanced plastic” are designed and intended for use in
military/aerospace applications or environments. Buyer acknowledges and agrees that any military or aerospace use of TI components
which have not been so designated is solely at the Buyer's risk, and that Buyer is solely responsible for compliance with all legal and
regulatory requirements in connection with such use.
TI has specifically designated certain components as meeting ISO/TS16949 requirements, mainly for automotive use. In any case of use of
non-designated products, TI will not be responsible for any failure to meet ISO/TS16949.
Products
Applications
Audio
www.ti.com/audio
Automotive and Transportation
www.ti.com/automotive
Amplifiers
amplifier.ti.com
Communications and Telecom
www.ti.com/communications
Data Converters
dataconverter.ti.com
Computers and Peripherals
www.ti.com/computers
DLP® Products
www.dlp.com
Consumer Electronics
www.ti.com/consumer-apps
DSP
dsp.ti.com
Energy and Lighting
www.ti.com/energy
Clocks and Timers
www.ti.com/clocks
Industrial
www.ti.com/industrial
Interface
interface.ti.com
Medical
www.ti.com/medical
Logic
logic.ti.com
Security
www.ti.com/security
Power Mgmt
power.ti.com
Space, Avionics and Defense
www.ti.com/space-avionics-defense
Microcontrollers
microcontroller.ti.com
Video and Imaging
www.ti.com/video
RFID
www.ti-rfid.com
OMAP Applications Processors
www.ti.com/omap
TI E2E Community
e2e.ti.com
Wireless Connectivity
www.ti.com/wirelessconnectivity
Mailing Address: Texas Instruments, Post Office Box 655303, Dallas, Texas 75265
Copyright © 2013, Texas Instruments Incorporated