Inductors and Chokes In Inductors and Chokes In Switch mode SuppliesSwitch mode Supplies

Thach, HungThach, Hung

12/06/0312/06/03

EE136EE136

TERMSTERMS

• Inductors – is reserved for wound Inductors – is reserved for wound components which DO NOT carry DC components which DO NOT carry DC current.current.

• Chokes – will be used for wound Chokes – will be used for wound components that carry a large DC components that carry a large DC bias current, with relatively small ac bias current, with relatively small ac ripple current. ripple current.

Design ApproachDesign Approach

• It will depend on the application. It will depend on the application.

• It is often a compromise, with It is often a compromise, with emphasis being placed on: emphasis being placed on:

• 1) Minimum cost 1) Minimum cost

• 2) Minimum size2) Minimum size

• 3) Minimum loss3) Minimum loss

Switch Mode ClassificationSwitch Mode Classification(Inductors)(Inductors)• Inductors will normally be confined to Inductors will normally be confined to

low pass filters. low pass filters.

• Their function is to prevent the Their function is to prevent the conduction of high frequency noise conduction of high frequency noise back into the supply lines. back into the supply lines.

• For this application, high core For this application, high core permeability would be an advantage.permeability would be an advantage.

ChokesChokes

• They will be found in high frequency They will be found in high frequency power output filters and continuous-power output filters and continuous-mode buck boost converter mode buck boost converter “transformers.”“transformers.”

• In these applications, low In these applications, low permeability and a low high-permeability and a low high-frequency core loss would be frequency core loss would be normally be considered an normally be considered an advantage. advantage.

ProblemsProblems

• To minimize the number of turns and To minimize the number of turns and copper loss, it might be assumed that a copper loss, it might be assumed that a high-permeability core material with a high-permeability core material with a low core loss would be the most low core loss would be the most desirable. desirable.

• In choke design, the large DC current In choke design, the large DC current component and the limited saturation component and the limited saturation flux density of real magnetic materials flux density of real magnetic materials force the selection of a low-permeability force the selection of a low-permeability material or introduction of an air gap in material or introduction of an air gap in the core. the core.

• As a result of low effective As a result of low effective permeability, more turns are needed permeability, more turns are needed to obtain the required inductance. to obtain the required inductance.

• So, in choke design, the desired low So, in choke design, the desired low copper and high efficiency are copper and high efficiency are compromised by the need to support compromised by the need to support a large DC current. a large DC current.

Simple InductorsSimple Inductors In power supply applications, pure In power supply applications, pure

inductors (those which do not carry a inductors (those which do not carry a DC component or a forced high-current DC component or a forced high-current ac component) are rare. ac component) are rare.

The design of these inductors is The design of these inductors is relatively easy (the inductance may be relatively easy (the inductance may be obtained by the Aobtained by the AL L value provided for value provided for the core, because no gap is required. the core, because no gap is required.

L = N² X AL = N² X ALL

Common-Mode Line Filter Common-Mode Line Filter InductorsInductors

Common-mode filter inductors have two Common-mode filter inductors have two isolated windings with the same number of isolated windings with the same number of turns. turns.

The 2 windings are connected so that the The 2 windings are connected so that the magnetic field that results from normal series-magnetic field that results from normal series-mode ac supply currents will cancel to zero.mode ac supply currents will cancel to zero.

The only inductance presented will be leakage The only inductance presented will be leakage inductance between the two windings. inductance between the two windings.

The low-frequency line current will not The low-frequency line current will not saturate the core, and a high permeability saturate the core, and a high permeability material may be used without the need for a material may be used without the need for a core air gap. core air gap.

Large Inductance can be obtained with a few Large Inductance can be obtained with a few turns. turns.

For common-mode noise (noise For common-mode noise (noise currents or voltages which appear on currents or voltages which appear on both lines at the same time with both lines at the same time with respect to the ground), the 2 respect to the ground), the 2 windings are in parallel and in phase, windings are in parallel and in phase, and a very high inductance is and a very high inductance is presented to common-mode presented to common-mode currents. currents.

Design ExampleDesign Example

In this example, it will be assumed that the In this example, it will be assumed that the maximum Common Mode Inductance is maximum Common Mode Inductance is required from a specified core size, using a required from a specified core size, using a high-permeability ferrite E core. high-permeability ferrite E core.

The effective DC or Low-frequency ac The effective DC or Low-frequency ac current in the core is zero as a result of current in the core is zero as a result of using 2 equally opposed and balanced using 2 equally opposed and balanced winding. winding.

Core loss is assumed to be negligible Core loss is assumed to be negligible because the design is to obtain the because the design is to obtain the maximum possible inductance at the maximum possible inductance at the working current from a particular core size. working current from a particular core size.

Core SizeCore Size

We want to select a core size that suits the We want to select a core size that suits the mechanical size requirement. mechanical size requirement.

Then calculate the area product (AP). The Then calculate the area product (AP). The area product is the product of the core area product is the product of the core area and the usable winding window area. area and the usable winding window area.

Refer to the core area product graph to Refer to the core area product graph to obtain the thermal resistance of the obtain the thermal resistance of the finished inductor. finished inductor.

AP = AAP = ACPCP X A X AWbWb cm^4cm^4

Core Area Product GraphCore Area Product Graph

Winding DissipationWinding Dissipation Now we need to calculate the Now we need to calculate the

permitted winding dissipation W that permitted winding dissipation W that will give an acceptable temperature will give an acceptable temperature rise T. rise T.

Then we can obtain the winding Then we can obtain the winding resistance Rresistance Rw w at the working (rms) at the working (rms) current I.current I.

Assuming zero core loss,Assuming zero core loss,

W = T / RW = T / Rth th WW

RRw w = W / l² = W / l²

From this permitted maximum From this permitted maximum resistance, the wire gauge, turns, resistance, the wire gauge, turns, and inductance can be established.and inductance can be established.

Establishing Wire Size, Turns, Establishing Wire Size, Turns, and Inductanceand Inductance

Many manufacturers provide Many manufacturers provide information on the resistance and information on the resistance and maximum number of turns of a fully maximum number of turns of a fully wound bobbin using various wire wound bobbin using various wire gauges. gauges.

The AL factors for the core are often The AL factors for the core are often provided, from which the inductance provided, from which the inductance can be calculated. can be calculated.

With balanced windings, there is no With balanced windings, there is no need for an air gap. need for an air gap.

A nomogram is used to, from which A nomogram is used to, from which the wire gauge, turns, and resistance the wire gauge, turns, and resistance of the wound component can be read of the wound component can be read directly. directly.

An inductor wound following the An inductor wound following the preceding steps provide the preceding steps provide the maximum inductance possible on the maximum inductance possible on the selected core size, at the maximum selected core size, at the maximum rated current and selected rated current and selected temperature rise. temperature rise.

Graphical Design of A Graphical Design of A Common-Mode-Line-Filter Common-Mode-Line-Filter Inductor (using a Ferrite E Inductor (using a Ferrite E

core)core)

AssumptionsAssumptions

1) EC35 Core is being used to provide the 1) EC35 Core is being used to provide the maximum inductance for a CML filter maximum inductance for a CML filter inductorinductor

2) Temperature rise does not exceed 302) Temperature rise does not exceed 30°C°C

3) Input Current is 5 A rms.3) Input Current is 5 A rms.

Nomogram for establishing wire size Nomogram for establishing wire size for chokes in ferrite material, as a for chokes in ferrite material, as a

function of turns and core sizefunction of turns and core size

The AP of the EC35 is 0.7 (when bobbin is The AP of the EC35 is 0.7 (when bobbin is used). used).

With AP = 0.7, the thermal resistance is With AP = 0.7, the thermal resistance is 2020°C/W.°C/W.

The dissipation for a temperature rise of 30°C The dissipation for a temperature rise of 30°C will be :will be :

Power = T / Rth = 30 / 20 = 1.5 WPower = T / Rth = 30 / 20 = 1.5 W

At a Current of 5 A rms, the maximum resistance will At a Current of 5 A rms, the maximum resistance will be related to power.be related to power.

P = IP = I²R²RR = 1.5 / 25 = 0.06 R = 1.5 / 25 = 0.06

By looking at the 2By looking at the 2ndnd nomogram, you can see that 0.06 nomogram, you can see that 0.06 will give you about 56 turns and wire gauge of will give you about 56 turns and wire gauge of about 17 (of AWG).about 17 (of AWG).

Note: In common-mode inductor, the winding will be Note: In common-mode inductor, the winding will be split into 2 equal parts. Hence, the EC35 bobbin split into 2 equal parts. Hence, the EC35 bobbin would be wound with 2 windings of 28 turns of #17 would be wound with 2 windings of 28 turns of #17 AWGAWG

Calculating Inductance (for common-mode inductors wound on

Ferrite E cores)

• In dual winding, common-mode inductors, the series-mode line frequency or DC magnetization force will cancel out. High permeability core may be used and a core gap is not required.

• For the previous example, AL value for the EC35 w/o an air gap is approximately 2000nH.

• The inductance for each 28-turn winding can be calculated as follows:

• L = N²X AL

• For the previous exampleL = 28²X 2000E-9 = 1.57 mH

Note: This graphical design approach also gives the maximum common-mode inductance that can be obtained from this core at 5 A for a temperature rise of 30°C.

CHOKESCHOKES

Inductors with DC Bias CurrentInductors with DC Bias Current

Brief ReviewBrief Review

Chokes (inductors which carry a large Chokes (inductors which carry a large component of DC current).component of DC current).

They are found in some form in all switch They are found in some form in all switch mode supplies. mode supplies.

Chokes range from small ferrite beads Chokes range from small ferrite beads used, for example, to profile the base drive used, for example, to profile the base drive currents of switching transistors, up to the currents of switching transistors, up to the very large high-current chokes used in very large high-current chokes used in power output filters. power output filters.

Design ConsiderationsDesign Considerations

1) Core Material1) Core Material 2) Core Design2) Core Design 3) Core Size3) Core Size 4) Winding Design4) Winding Design

Note: Since this subject is very broad, this Note: Since this subject is very broad, this discussion will be confined to those types of discussion will be confined to those types of chokes most often used in high-frequency switch chokes most often used in high-frequency switch mode applications. mode applications.

Core MaterialCore Material

The Core material is chosen to suit The Core material is chosen to suit the following conditions:the following conditions:

1) The Operating Frequency1) The Operating Frequency

2) Ratio of DC to ac Current2) Ratio of DC to ac Current

3) Inductance3) Inductance

4) The mechanical requirements4) The mechanical requirements

Core SizeCore Size

Often the most difficult choice is the Often the most difficult choice is the core size and configuration. core size and configuration.

There are many different core topologies There are many different core topologies that exist, so it may be difficult to that exist, so it may be difficult to decide which would be the optimum decide which would be the optimum choice for a particular application. choice for a particular application.

The Area Product (AP) tends to be a The Area Product (AP) tends to be a reasonable constant for all core reasonable constant for all core topologies of the same general power topologies of the same general power rating, and this can be used for the core rating, and this can be used for the core size.size.

Area ProductArea Product

It is the product of the winding It is the product of the winding window area and the core center window area and the core center pole area.pole area.

In general, AP = Aw * Ac cm ^ 4In general, AP = Aw * Ac cm ^ 4

Temperature RiseTemperature Rise

The temperature rise of the wound The temperature rise of the wound component in free air cooling component in free air cooling conditions will depend on the total conditions will depend on the total loss in the wound component and the loss in the wound component and the comment's surface area. comment's surface area.

The actual temperature rise, The actual temperature rise, T, T, that that may be expected from a particular may be expected from a particular core size AP is given by :core size AP is given by :

T = P * RtT = P * Rt T – temperature rise, °CT – temperature rise, °C P – the total dissipation, WP – the total dissipation, W Rt – thermal Resistance, °C/WRt – thermal Resistance, °C/W

Note: In choke design, the loss P will Note: In choke design, the loss P will be mainly copper loss. Core Losses be mainly copper loss. Core Losses are small in most cases, and may be are small in most cases, and may be neglected. neglected.

Core Air GapsCore Air Gaps

If considerable DC currents flow in the If considerable DC currents flow in the choke, the use of gapped E or C cores choke, the use of gapped E or C cores may be considered. may be considered.

Since chokes will normally be required Since chokes will normally be required to support the DC component w/o to support the DC component w/o saturation, relatively large air gaps are saturation, relatively large air gaps are used, and the effective permeability, used, and the effective permeability, irrespective of the material chosen, is irrespective of the material chosen, is usually very low – around 10 and 300usually very low – around 10 and 300

Ferrite material saturates at lower flux Ferrite material saturates at lower flux density than iron, even when gapped. density than iron, even when gapped.

To prevent saturation, a large gap To prevent saturation, a large gap must be used in the core, resulting in must be used in the core, resulting in a lower effective permeability and a lower effective permeability and giving lower inductance. giving lower inductance.

The higher saturating flux density of The higher saturating flux density of iron core permits a smaller gap, giving iron core permits a smaller gap, giving a larger permeability for the same DC a larger permeability for the same DC bias conditions; hence inductance is bias conditions; hence inductance is greater, and the ripple current will be greater, and the ripple current will be smaller. smaller.

A further advantage of the gapped E A further advantage of the gapped E or C core is that the effective or C core is that the effective permeability can be optimized for the permeability can be optimized for the application by adjusting the gap size application by adjusting the gap size for the most effective performance. for the most effective performance.

ConclusionConclusion

The core size depends on the total loss The core size depends on the total loss and the permitted temperature rise. and the permitted temperature rise.

The copper loss depends on the DC The copper loss depends on the DC current, turns, and wire size.current, turns, and wire size.

The core loss, and hence the choice of The core loss, and hence the choice of material, depends on the ac volt-material, depends on the ac volt-seconds that the choke must withstand, seconds that the choke must withstand, that is, the flux density swing that is, the flux density swing B and B and the operating frequency. the operating frequency.

Design Examples of a Design Examples of a Gapped Ferrite E-core Gapped Ferrite E-core

ChokeChoke

Using an Empirical MethodUsing an Empirical Method

AssumptionsAssumptions The choke is required to support a large DC The choke is required to support a large DC

current with considerable high-frequency current with considerable high-frequency ripple currentripple current

A low-core loss material is usedA low-core loss material is used An air gap is requiredAn air gap is required The maximum core size is defined by the The maximum core size is defined by the

mechanical rather than the ideal electrical mechanical rather than the ideal electrical needs. needs.

Note: Typical applications would be an output Note: Typical applications would be an output filter inductor for a high-frequency forward filter inductor for a high-frequency forward converter. DC current is 10 A and ripple converter. DC current is 10 A and ripple current does not excced 3 A at 100Khzcurrent does not excced 3 A at 100Khz

1) Select core and bobbin size (as 1) Select core and bobbin size (as defined by mechanical needs), and defined by mechanical needs), and completely fill the bobbin with a completely fill the bobbin with a gauge of wire that will give gauge of wire that will give acceptable Power loss and hence acceptable Power loss and hence acceptable temperature rise. acceptable temperature rise.

2) Assemble core and bobbin, 2) Assemble core and bobbin, allowing adequate air gap. allowing adequate air gap.

3) Fit the choke in the power filter 3) Fit the choke in the power filter position in the supply and observe position in the supply and observe the choke ripple current waveform.the choke ripple current waveform.

4) Adjust the air gap under 4) Adjust the air gap under maximum load and input voltage maximum load and input voltage conditions until a minimum ripple conditions until a minimum ripple current is observed. current is observed.

Note: By this, maximum dynamic Note: By this, maximum dynamic inductance has now been obtained. inductance has now been obtained.

Presentation ConclusionPresentation Conclusion

This presentation was about mainly This presentation was about mainly designing basic Inductors and a brief designing basic Inductors and a brief explanations about how basic chokes explanations about how basic chokes are designed. are designed.

I would like to thank Dongsheng I would like to thank Dongsheng Zhou, Ph.D.Zhou, Ph.D.

for giving us the opportunity to learn for giving us the opportunity to learn about something interesting. about something interesting.

ReferencesReferences

““Switchmode Power Supply Switchmode Power Supply Handbook”Handbook” 2nd edition authored by 2nd edition authored by Keith Billings published by McGraw Keith Billings published by McGraw Hall. (chapter 3.1)Hall. (chapter 3.1)