Eddy Current Losses in TransformerWindings and Circuit Wiring

Lloyd H. Dixon, Jr.

energy is stored in air gaps, insulation betweenconductors, and within the conductors, whererelative permeability JLr is essentially 1.0 andconstant. The energy density then becomes:

w = \BH = \'tJ.(j/I2 J/m3

where ILO is the absolute permeability of freespace ( =47r .10"7 in S.I. units). Total energy W(Joules) is obtained by integrating the energydensity over the entire volume, v, of the field:

W = \ILO f H2dv Joules

Within typical transformers and inductors, themagnetic energy is almost always confined toregions where the field intensity H is relativelyconstant and quite predictable. This often oc-curs in circuit wiring, as well. In these cases:

W = \ILO H2A. e Joules (2)

and from (1), He = NI. Substituting for H:

W = \ILO ~fA/e Joules (3)

whereA is the cross-section area (m1 of theregion normal to the flux, and e is the lengthof the region in meters (and the effectivelength of the field).

4. Circuit inductance: Inductance is a meas-ure of an electrical circuit's ability to storemagnetic energy. Equating the energy stored inthe field from (3) with the same energy incircuit terms:

W = \Lf = \ILO N2fA/e

L = ILO N2A/e (4)

Skin EffectFigure 1 shows the magnetic field (flux lines)

in and around a conductor carrying dc or lowfrequency current I. The field is radially sym-metrical, as shown, only if the return currentwith its associated field is at a great distance.

At low frequency, the energy in the magneticfield is trivial compared to the energy loss inthe resistance of the wire. Hence the current

IntroductionAs switching power supply operating fre-

quencies increase, eddy current losses andparasitic inductances can greatly impair circuitperformance. These high frequency effects arecaused by the magnetic field resulting fromcurrent flow in transformer windings and circuit

wiring.This paper is intended to provide insight into

these phenomena so that improved high fre-quency. performance can be achieved. Amongother things, it explains (I) why eddy currentlosses increase so dramatically with more wind-ing layers, (2) why parallelling thin strips does-n't work, (3) how passive conductors (Faradayshields and C. T .windings) have high losses,and (4) why increasing conductor surface areawill actually worsen losses and parasitic induc-tance if the configuration is not correct.

Basic PrinciplesThe following principles are used in the

development of this topic and are presentedhere as a review of basic magnetics.

1. Ampere's Law: The total magneto-motiveforce along any closed path is equal to the totalcurrent enclosed by that path:

F = iHde = I, = NI Amps (1)

where F is the total magneto-motive force(in Amperes) along a path of length e (m), His field intensity (Aim), and I, is the totalcurrent through all turns enclosed by the path.

2. Conservation or energy: At any momentof time, the current within the conductors andthe magnetic field are distributed so as tominimize the energy taken from the source.

3. Energy content or the field: The magneticfield is energy. The energy density at any pointin the field is:

w = f HdB Jouleslm3

where B is the flux density (Tesla). Inswitching power supplies, almost all magnetic

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reduced, so the resistance at high frequency( and resulting losses) can be many timesgJ:eater than at low freQuencv.

Fig. 1 -IsQ/ated CQnductQr at LQW Frequency

distributes itself uniformly throughout the wireso as to minimize the resistance loss and thetotal energy expended.

Around any closed path outside the wire(and inside the return current), magneto-motiveforce F is constant and equal to total current I.But field intensity H varies inversely with ther,adial distance, because constant F is appliedacross an increasing t ( =2m).

Within the conductor, F at any radius mustequal the enclosed current at that radius,therefore F is proportional to r2.

Fig. 2 -Eddy Current at High Frequency

At high frequency: Figure 2 is a super-position model that explains what happenswhen the frequency rises. The dash lines repre-sent the uniform low frequency current distri-bution, as seen in Fig. I. When this currentchanges rapidly, as it will at high frequency, theflux within the wire must also change rapidly.The changing flux induces a voltage loop, oreddy, as shown by the solid lines near the wiresurface. Since this induced voltage is within aconductor, it causes an eddy current coincidentwith the voltage. Note how this eddy currentreinforces the main current flow at the surface,but opposes it toward the center of the wire.

The result is that as frequency rises, currentdensity increases at the conductor surface anddecreases toward zero at the center. as shownin Fig. 3. The current tails off exponentiallywithin the conductor. The portion of theconductor that is actually carrying current is

Fig. 3 -High Frequency Cu"ent Distribution

Penetration depth: Penetration or skindepth, OPEN , is defmed as the distance fromthe surface to where the current density is l/etimes the surface current density ( e is the

naturallog base)[l]:OPEN = [ p/(1TIJ.£) ]112 m (5)

where p is resistivity. For copper at 100°C,p = 2.3.10-6 n-cm, IJ.= IJ.o = 41T.10°7, and:

OPEN = 7.5/(£)112 cm (6)

From (6), OPEN = .024 cm at 100 kHz, or.0075 cm at 1 MHz.

Eqs. (5) and (6) are accurate for a flat con-ductor surface, or when the radius of curvatureis much greater than the penetration depth.

Although the current density tails off expo-nentially from the surface, the high frequencyresistance (and loss) is the same as if the cur-rent density were constant from the surface tothe penetration depth, then went abruptly tozero as shown on the right hand side of Fig. 3.This equivalent rectangular distribution iseasier to apply.

Equivalent circuit model: Another way oflooking at the high frequency effects in trans-former windings and circuit wiring is throughthe use of an equivalent electrical circuitmodel. This approach is probably easier for acircuit designer to relate to.

Figure 4 is the equivalent circuit of theisolated conductor of Figs. 1 to 3. With currentI flowin, through the wire, L% accounts for theenergy ~L%I2 stored in the external magneticfield. L% is the inductance of the wire at high

frequencies.Point A represents the outer surface of the

conductor, while B is at the center. R; is the

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Fig. 4 -Conductor Equivalent Circuit

resistance, distributed through the wire fromsurface to center. Think of the wire as dividedinto many concentric cylinders of equal crosssection area. The Ri elements shown in thedrawing would correspond to the equal resis-tance of each of the cylinders. Likewise, theinternal inductance L; accounts for the magnet-ic energy distributed through the cylindricalsections. The energy stored in each sectiondepends on the cumulative current flowingthrough the elements to the right of thatsection in the equivalent circuit.

Note that external inductance Lx of the wire(or the leakage inductance of a winding) limitsthe maximum di/dt through the wire, depend-ing on the source compliance voltage, nomatter haw fast the switch turns on.

The time domain: If a rapidly rising currentis applied to the wire, the voltage across thewire is quite large, mostly across Lx. Internalinductance Li blocks the current from the wireinterior, forcing it to flow at the surfacethrough the left-most resistance element, evenafter the current has reached its final value andthe voltage across Lx collapses. Although theenergy demand of Lx is satisfied, the voltageacross the wire is still quite large becausecurrent has not penetrated significantly into thewire and must flow through the high resistanceof the limited cross-section area at the surface.Additional source energy is then mostly dissi-pated in the resistance of this surface layer.

The voltage across this Ri element at thesurface is impressed across the adjacent Li ele-ments toward the center of the wire, causingthe current in L; near the surface to rise.Current cannot penetrate without a field beinggenerated within the conductor, and thisrequires energy. As time goes on, conductionpropagates from the surface toward the center(at B in the equivalent circuit), storing energyin L;. More of the resistive elements conduct,

lowering the total resistance and reducing theenergy going into losses. Finally, conduction isuniform throughout the wire, no further energygoes into the magnetic fields external orinternal to the wire, and a small amount ofenergy continues to be dissipated in Ri overtime.

Note that the concept of skin depth has nomeaning in the time domain.

The frequency domain: Referring again toFig. 4 with a sine wave current applied to theterminals, it is apparent that at low frequencies,the reactance of internal inductance Li is neg-ligible compared to Ri. Current flow is uniformthroughout the wire and resistance is minimum.But at high frequency, current flow will begreatest at the surface (A), tailing off exponen-tially toward the center (B).

Penetration depth (skin depth) is clearlyrelevant in the frequency domain. At anyfrequency, the penetration depth from Eq. (5)or (6) reveals the percentage of the wire areathat is effectively conducting, and thus the ratioof dc resistance to ac resistance at that

frequency.Although the current waveforms encountered

in most switching power supplies are notsinusoidal, most papers dealing with the designof high frequency transformer windings use asinusoidal approach based on work done byDowell in 1966.[2] Some authors use Fourieranalysis to extend the sinusoidal method tonon-sinusoidal waveforms.

Proximity EffectUp to this point a single isolated conductor

has been considered. Its magnetic field extendsradially in all directions, and conduction occursacross the entire surface.

When another conductor is brought intoclose proximity to the first, their fields addvectorially. Field intensity is no longer uniformaround the conductor surfaces, so high frequen-cy current flow will not be uniform.

For example, if the round wire of Fig. 1 isclose to another wire carrying an equal currentin the opposite direction (the return currentpath?), the fields will be additive between thetwo wires and oppose and cancel on the out-sides. As a result, high frequency current flowis concentrated on the wire surfaces facing eachother, where the field intensity is greatest, with

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little or no current on the outside surfaceswhere the field is low. This pattern arrangesitself so as to minimize the energy utilized,hence inductance is minimized. As the wiresare brought closer together, cancellation ismore complete. The concentrated field volumedecreases, so the inductance is reduced.

Circuit wiring: The field and current distri-bution with round wires is not easy to compute.A sinlpler and more practical example is givenin Figure 5. The two flat parallel strips shownare actually the best way to implement highfrequency wiring, minimizing the wiring induc-tance and eddy current losses. These stripscould be two wide traces on opposite sides ofa printed circuit board. (Don't use point-to-point wiring. It is much more important tocollapse the loop and keep the outgoing andreturn conductors as intimate as possible, evenif the wiring distances are greater .)

I---t--"i

1+ + + + +11-

1. iT

length, divided equally between each of the twoconductors. If one conductor is much widerthan the other, such as a strip vs. a groundplane, most of the inductance calculated in (6)is in series with the narrower conductor. Thisis good for keeping down noise in the groundreturns.Note that current penetration is from one sideonly --the side where the field is. This meansthat a strip thicker than the penetration depthis not fully utilized. The equivalent circuitmodel of Fig. 4 still applies, with A at thesurface adjacent to the field. But B becomesthe opposite side of the strip, not the center,since there is no penetration from the side withno field.

Bad practice: Figures 6 and 7 show what notto do for circuit wiring (unless you want highinductance and eddy current losses). Althoughthese strips have large surface areas, proximityeffects in these configurations result in verylittle surface actually utilized. Remember thatthe field is concentrated directly between thetwo conductors so as to minimize the stored

energy.In Fig. 6 this results in current flow only at

the edges facing each other. Also, because theconcentrated field region is short, the energydensity is very high, and the inductance isseveral times greater than in Fig. 5.

The Fig. 7 configuration is not quite as badas Fig. 6 because the current does spread outsomewhat in one of the two conductors, but itis still many times worse than the proper con-figuration in Fig. 5. The message is: large

1 !1 11 1

Fig. 6 -Bad Wiring Practice -Side by Side

[~

Fig. 5 -Circuit Wiring -Flat Parallel Strip

The + signs indicate current flow into theupper strip, the. indicates current out of thelower strip. Between the strips, the magneticfield is high and uniform, so the current isspread out uniformly on the inner surfaces. Onthe outside of the strips the field is very low, sothe current is almost zero. This results in theminimum possible energy storage (and wiringinductance) for this configuration. If thebreadth of the strip, t, is much greater thanthe separation, w, the energy is almost entirelycontained between the two strips. Then t andware the length and width of the field, and canbe used to calculate the inductance. Convertin~Eq. (4) to cm and with N = 1 turn, the induc-tance per centimeter length of the 2-conductorstrip is:

L = 12.5 wit oHlcm (7)

If the strips have a breadth of 1 cm and areseparated by 0.1 cm, the combined inductanceof the pair is only 1.25 oH for each centimeter

Fig. 7 -Bad Wiring Practice -Right Angled

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surface area doesn't improve high frequencyperformance if the configuration is wrong.

Inductor Windings: Figure 8 is a simpleinductor. The winding consists of 4 turns in a

single layer. Assuming a current of 1 Amperethrough the winding, the total magneto-motiveforce F = NI along any path linking the 4 turns

is 4 Ampere-turns. The field is quite linear

across the length of the window because of theaddition of the fields from the individual wiresin this linear array. The winding could havebeen a flat strip carrying 4 Amps with the sameresult.

I~

-

Fig. 8 -Inductor Winding

Fig. 9 -Transfonner Windings

is neglected). So if the secondary load currentis 4 A through 1 turn, the primary currentthrough 4 turns must be 1 Amp. The fieldstend to cancel not only outside both windings,but in the center of the two windings as well.Whatever field might remain is shorted out bythe high permeability core which has no gap.Thus the field generated by the current in thewindings, F = 4 A, exists only between the

windings: So at high frequency, current flow ison the outside of the inner winding and on theinside of the outer winding, adjacent to thefield.

Multiple Layer WindingsFigure 10 shows a transformer with multi-

layer windings and its associated low frequencymmf (F) and energy density diagrams. One halfof the core and windings are depicted. At lowfrequency, current (not shown) is uniformlydistributed through all conductors, because theyare much thinner than the penetration depth.

Without the ferrite core, the field outside ofthe winding would have been weak because ofcancellation, but with the high permeabilitycore, the external field is completely shortedout. This means the entire field, F = NI iscontained across the window inside the wind-ing. Field intensity, H equals NI/t. In thecenter, the entire field is compressed across thesmall air gap. Field intensity (H, = NI/t.) istherefore much greater within the gap, so theenergy stored in the gap (using Eq. 2 or 3) ismuch greater than the energy in the much larg-er window.

At high frequency, current flow is concentra-ted on the inner surface of the coil, adjacent tothe magnetic field. The field outside the coil isnegligible, so no current flows on the outersurface.

Transformer Windings: Figure 9 shows atransformer with a four turn single layerprimary and a 1 turn single layer copper stripsecondary. In any transformer, the sum of theAmpere-turns in all windings must equal zero(except for a small magnetizing current which

F"=Ni~L

JL~LFig. 10- Mu/tip/e Layer Winding

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interleaving, as shoWn in Figure 11.With interleaving, each winding is essentially

divided into two or more sections, as shown oneach side of the dotted line in rIg. 11. Theprimary now has two sections a and b, eachwith 1 layer of 4 turns. The secondary is alsodivided down the middle into two sections of 1~ layers, 1 turn per layer. (This is why halflayers are included in eddy current loss curves.)Note that the 2 amperes through the 1 ~ turnsof secondary section (a) cancels the 0.75 A, 4turns of primary section (a), and the field goesthrough zero in the middle of the centersecondary turn at the dotted line. F builds upto only half the peak value compared to Fig.10, and reverses direction between alternatewinding sections. It will be shown that thereduced field causes a great reduction in eddycurrent losses.

Because of the reduced field, the total ener-gy under the energy density curve in Fig. 11 isonly 1/4 the total energy in Fig. 10. Thus,interleaving reduces the leakage inductance bya factor of 4!

Multiple layers at high frequency: Figure 12is an enlarged section of Fig. 10 but at a highf(equency where the penetration depth is 20%of the secondary winding strip thickness. Thefield strength and current density are the sameas at low frequency in the spaces between theconductors. But within the conductors, currentdensity, magnetic field and energy density allfall off rapidly moving in from the surface.(Dash lines show low freQuency distributions.

The primary winding has 8 turns arranged intwo 4-turn layers, while the secondary has 3turns of copper strip in 3 layers. With 2 Ampload current, the secondary has 3T.2A=6 A-T.The primary must also have 6 A- T, so primarycurrent is 0.75 A.

As shown in the mmf diagram below thecore in Fig. 10, there is no field outside of theprimary or inside the secondary, but starting atthe outside of the primary and moving towardthe center, the field rises to its maximum valuebetween the two windings. With uniform cur-rent distribution at low frequency, note howthe field builds uniformly within each conductoraccording to Faraday's law, staying constantbetween the conductors. The energy density inthe field goes up with the square of the fieldstrength, as shown below the mmf diagram.The area under the energy density curve is thetotal leakage inductance energy stored in andbetween the windings.

So multiple layers cause the field to build.At high frequencies, it will be shown that theeddy current losses go up exponentially as thenumber of layers is increased. The number oflayers should be kept to a minimum by using acore with a long narrow window toaccommodate all the turns in fewer layers (thisalso causes a dramatic reduction in leakageinductance). The window shape illustrated inFig. 10 is far from optimum.

Interleaved Windings: Another way to re-duce the effective number of layers is to breakup the winding into smaller sections through

f=N~~~w l ~ : -",,""",,-1

Fig. 11 -Inter/eaved Windings Fig. 12 -Multiple Layers at High Frequency

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The heavier lines showing the high frequencydistributions are approximations based on thecurrent distribution ending abruptly at thepenetration depth. Actually, the slopes aresteeper at the surfaces, and tail off within the

wire.)Notice that the total energy under the energydensity curve is about half the amount at lowfrequency. This means the leakage inductancehas decreased at high frequency, but onlybecause the energy within the conductors ispractically eliminated. This is not a verypractical way to reduce leakage inductance --the penalty is a huge increase in eddy currentlosses.

With the penetration depth 20% of the stripthickness, the ac resistance might be expectedto be 5 times the dc resistance. But with threelayers building up the field, the ac resistance isactually 32 times greater!

Why multiple layers cause high losses:Figure 13 is an even more enlarged view of the3 secondary layers of Fig. 12, together withtheir high frequency mmf diagram. Assume acurrent of I Ampere through the 3 layerwinding. There is only one turn (strip) per

layer.To the right of layer SI, the mmf is 0. At

the left of SI, F= I A-T. The I Amp currentin SI is crowded into the 20% penetrationdepth adjacent to the field. Field F also cannotpenetrate more than 20% into SI. This I A-T

9/3=

field exists only between Sl and S2. F cannotpenetrate S2, either. It must terminate on theright side of S2, but it cannot just magically

disappear.According to Faraday's law, for the mmf to

be zero in the center of S2, the enclosedcurrent must be zero. This requires a currentof 1 Amp on the right surface of S2 in theopposite direction to nonnal current .flow tocancel the 1 Amp in Sl. Then, to achieve the 2A-T field to the left of S2 requires a surfacecurrent of 2 Amps! (The net current in S2 isstill1 Amp.)

The 2 A- T field must be terminated at theright side of S3 with 2 A reverse current flow.Then 3 Amps must flow on the S3 left surfaceto support the 3 A- T field and to conform tothe net 1 Amp through the winding.

If the current in each layer were just the 1Amp, limited in penetration to 20% of the con-ductor thickness, the ac to dc resistance ratio,FR, would be 5:1. But the surface currents insuccessive layers become much larger, as dis-cussed above. The tabulation above the conduc-tors in Fig. 13 gives the current at each surfaceand the current squared, which indicates therelative power loss at each surface. The twosurfaces of S2 together dissipate 1 + 4 = 5 timesas much as Sl, while dissipation in S3 is4+9=13 times Sl!

The average resistance of the 3 layers is(9+4+4+ 1 + 1)/3 = 19/3 = 6.333 times theresistance of layer Sl. Since the ac resistance ofSl is already 5 times the dc resistance (becauseof the 20% penetration depth), the ratio ofaverage ac resistance to the dc resistance, FR,is 31.67 : 1. Hardly negligible.

Referring to S3 with 3A and -2A at itssurfaces, if conductor thickness is decreased orif frequency is decreased to improve penetra-tion, the penetration "tails" of these opposingcurrents will reach across and partially cancel.When DpEN is much larger than the conductorthickness, cancellation is complete, conductorcurrent is 1 Amp uniformly distributed, andFR = 1.

Although each layer was a 1 turn stripcarrying 1 Amp in this illustration, each layercould have been 10 turns carrying 0.1 Ampswith the same results.

Fig. ]3 -Surface Currents

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1 -129 9

3 -3

9 4 4 1

3 -2 2 -1

Paralleled windings: When ac resistance ofa strip secondary winding is too high becausethe required strip thickness is too great, it istempting to simply subdivide it into severalthinner strips, insulated from each other. Thisdoesn't work --the parallel combination willhave the same losses as an equivalent solidstrip. This is because the individual thin stripsoccupy different positions in the field, causingeddy currents to circulate between the outerand innermost strips where they are connectedin parallel at their ends, similar to whathappens in a single solid strip.

Conductors can be successfully paralleledonly when they experience the same field,averaged along their length:1. Wires in the same layer can be paralleled, as

long as they progress together from onelayer to the next.

2. Litz wire --fme wires woven or twisted insuch a manner that they successively occupythe same positions in the field.

3. Portions of a winding at comparable fieldlevels in different interleaved sections can beparalleled. For example the two four-turnprin1ary sections in Fig. 11 could be paral-leled. The field must remain apportionedequally between in the two sections or muchmore energy would be required. If the sec-ondary in Fig. 11 were a single solid turn, itcould be divided into two thinner paralleledturns.

Calculating ac resistance: It has been shownthat it is not difficult to calculate FR when theconductor thickness is much greater than thepenetration depth. It is also easy when theconductor thickness is much less than DPEN --FR = 1. But the condition of greatest interestto the transformer designer is when the con-ductor thickness is in the same range as' DPEN ,and here the calculations are quite difficult.

Dowell solved the problem for sinusoidalwaveforms in his 1966 paper .[2] The curves inFigure 15 are derived from Dowell's work. Thevertical scale is FR, the ratio of Rac/Rdc. Thehorizontal scale, a, is the ratio of the effectiveconductor height, or layer thickness, to thepenetration depth, DPEN .For strip or foil wind-ings, the layer thickness is the strip thickness.For round wires touching each other in thelayer, the effective layer thickness is 0.83 times

Fig. 14 -Passive Winding Losses

Passive layers: A passive layer is any con-ductor layer that is not actively "working" bycarrying net useful current. Faraday shields andthe non-conducting side of center-tapped wind-ings are examples of passive layers.Figure 14 shows what happens if a Faradayshield is inserted in the 3 A-T field betweenthe secondary winding of Fig. 13 and theprimary (off to the left). If the Faraday shieldthickness is much greater than DPEN , 3 Ampsmust flow on each surface because the fieldcannot penetrate. At each surface, 12 is 9 Ampssquared, and both surfaces together dissipate18 times as much as SI, or almost as much asall three secondary turns combined.

Faraday shields are always located where thefield is highest. Their thickness should be lessthan 1/3 of OPEN keep the loss to an accept-able level.

For another example, consider a center-tapped secondary with sides A and B each sideof the center-tap. If A is physically between Band the primary, A is a passive conductor inthe high field region when B is conducting, butB is outside the field when A conducts. Theadditional passive dissipation in A will probableexceed the active dissipation in either A or B.This is one reason that single-ended trans-formers overtake or even surpass push-pu11 andhalf-bridge versions at frequencies above a fewhundred kHz.

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All Frequencies

So

1./

/

~

u I-0 IOC 1~" IUOOC 5

,

~.25

substituted for theoriginal wire (takingcare to handle thisproperly), there willbe 40 wires, 20 perlayer, 2 layers deep.0 is now 2. Entering

15 at 0=2 and 2-" II layers, Fa is 5.2 --it

I I went up! The reasonthis is the wire

II size is still too large~ I for effective penetra- D£ tion and cancellation

of the eddy currents,

and the Dumber oflayers has been

II doubled with the5 a ~ I~ye, th;Ckn'ess/DPen .~I extra eddy current5 1 2 4 10 surfaces this gener-

Q = layer thickness/Open ates.

.Subdividing againFIg. 15 -Eddy Cu"ent Losses -RAc/Roc into 16 parallel wires

the wire diaDleter. For round wires spaced of 1/4 the original diameter there are 160 totalapart in a lal5r, the effective layer thickness is wires, 40 per layer, 4 layers deep. 0 is 1 and.83 .d .(d/s) , where d is wire diaDleter and s Fa is down to 2.8.is the center-to-center spacing of the wires. A third subdivision to 64 parallel wires with

Referring to the calculation of Fa = 31.67 1/8 the original diaDleter results in 640 totalon page 7, enter Fig. 15 with 0~5 and 3 layers. wires, 80 per layer, 8 layers. 0 is 0.5 and FaThe resulting Fa value agrees. fmally reaches 1.5.

At the extreme right of Fig. 15 is the region Non-sinusoidal wavefonns: Venkatramen[3]where the conductor thickness is much greater and Carsten[4] have applied Dowell's sine wavethen DPEN and Fa is very large. The curves are solution to various non-sinusoidal waveformsparallel and have a + 1 slope. On the extreme more relevant to switching power supplies. Thisleft, conductor thickness is much less than is done by taking the Fourier components ofDPEN and Fa approaches 1.0. In the center of the current waveform, then using Dowell'sthe graph, the curv,es plunge downward as 0 approach to calculate the loss for eachgets smaller. An Fa of 1.5 is a good goal to harmonic and adding the losses. They have alsoachieve. With a much higher value, losses hurt redefmed the way the data is presented in antoo much. To go much below 1.5 is past the attempt to make it more useful.point of diminishing returns, requires much Their curves show that as the pulse widthfmer conductor sizes. Achieving Fa of 1.5 Jtarrows, the effective ac resistance goes uprequires a 0 value ranging from 1.6 with 1 because the higher frequency harmonics arelayer, to 0.4 with 10 layers. more important. But the worst losses are not at

Starting with a conductor thickness much narrow pulse widths. In most switchinggreater than DpEN and subdividing into smaller supplies, the peak pulse current is constant (atconductors usually makes Fa worse before it full load). The high frequency harmonics andgets better. For example, assume a single layer losses are much the same regardless of pulsewinding of 10 close spaced turns, and a Q of 4. width, but the total rms, the dc and lowFa from the Fig. 15 is 3.8 --not good enough. frequency components get much worse as theIf four parallel wires of half the diameter are pulse width widens. Worst case for total copper

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losses is probably near a duty ratio of 0.5.In applications where current pulse widths in

the vicinity of 0.5 duty ratio are the worst caseconditions for copper losses, a shortcut methodto achieve satisfactory results is to design thewinding for an FR of 1.5 at the fundamental ofthe current waveform, then allow an extra 30-50% for additional losses due to the high

frequency components.

References

[I] D. G. Fink et al, Electronics Engineer'sHandbook, McGraw-Hill, 1975

[2] P. L. Dowell, "Effects of Eddy Currents inTransformer Windings," Proceedings IEE(UK), Vol. 113, No.8, August, 1966, pp.1387-1394.

[3] P. S. Venkatramen, "Winding Eddy CurrentLosses in Switch Mode Power Transform-ers Due to Rectangular Wave Currents,"Proceedings of Powercon 11, 1984, Sec. AI.

[4] B. Carsten, "High Frequency ConductorLosses in Switchmode Magnetics," HighFrequency Power Convener Conference,1986, pp. 155-176

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Appendix I Litz Wire

voids and poor copper utilization. Litz wire forpower applications is usually made with a fewwires twisted together in a strand and a few ofthese strands twisted into bigger strands, etc.The amount of twist required is moderate --

not enough to significantly increase the stranddiameter or the length of the individual wires.

Consider a bundle of seven wires twistedinto a strand, with one wire in the center. Thepacking structure is hexagonal --as efficient aspossible, with an outer surface which is almostcircular .The amount of copper in this strand is.778 of a single solid wire with the same max.diameter as the strand. The outer six wiresrotationally occupy each of the outside posi-tions, but the center wire is fIXed in its location.ac losses are actually improved by eliminatingthe seventh, inner wire. (In practice, it shouldbe replaced by a non-conductive filler to main-tain the shape of the strand.) This reduces thewinding area utilization factor to .667 of theequivalent solid wire.

Even solid wire does not achieve 100% util-ization of the winding area. The bottoms of:thewires in each layer ride diagonally across thetops of the wires in the layer below, so theycannot pack down into the valleys between thewires below, except in a limited and unpredict-able way. This means that round wire occupiesthe area of a square with sides equal to wirediameter, hence the utilization is pi/4, or .785.If a six-wire strand described above is substi-tuted for the solid wire, overall utilization isfurther reduced to. 785. .667 = .524.

Table I shows the utilization factor of 3 to6-wire strands. Although the utilization factorsare quite similar, it will be shown that they donot perform equally well in achieving a multi-strand cable with a large number of wires.

TABLE IUtilization F.ctor or Single Str.nds

NlI1Der of Wires:

Utilization Factor:

3

.65

4 5 6

.686 .685 .667

In switching power supply transformers, if theconductor thickness is similar to or greaterthan penetration (skin) depth at the operatingfrequency, ac current flows in only a portionof each conductor, resulting in high ac losses.This effect is magnified exponentially the morelayers there are in the winding. To bring theac losses back down to an acceptable level, theconductor thickness must be reduced.

Thin strip with a width equal to the windingwidth is often used, especially for low voltage,high current windings with few turns and largeconductor area. Each turn is a layer and eachturn must be insulated from the others. Stripscannot be subdivided into several paralleledthinner strips unless the individual strips are indifferent winding sections, otherwise unequalinduced voltages will cause large eddy currentsto circulate from one strip to another andlosses will be high.

When strip or foil is not appropriate for awinding, the conductor can be divided intomultiple strands of fine wire which are thenconnected in parallel at the terminations of the

winding.All wires in the group must be individually

insulated and must encompass the identical fluxto avoid eddy currents circulating from onewire to another through their terminating inter-connections. If only a few wires in parallel areto be used, they can be laid together side-by-side (as though they were tied together as a flatstrip). Each wire must have exact/y the samenumber of turns as the other wires within each/ayer, to avoid cross-circulating eddy currents.This is not practical when large numbers offme wires must be used.

Another solution is to interweave or twistthe wires together in such a manner that eachwire moves within the group to successivelyoccupy each level within the field. But whenthis is done properly, voids are introduced,resulting in poor utilization of the availablewinding area compared with closely packed(untwisted) conductors.

Low power, high frequency Utz wire isusually woven from very fme wires, but thewoven structure results in a large percentage of

R2-1

available breadth and height, and improves theutilization factor. by eliminating one level of

twisting.For example, assume 256 wires of a given

diameter are required. This could be achievedin one Level 4 cable using 4 strands/twist and4 twist levels. The utilization is .22 comparedwith solid wire the same diameter as the cable,and .22°.785 = .17 of the winding space iscopper. However if 4 Level 3 cables are used inparallel, the utilization is .32 compared to solidwire, and .32 ° .785 = .25 of the occupied wind-ing area. Remember that each of the paralleledcables must have the same number of turns ineach layer.

Insulation on the wires further reduces theutilization, especially with fine wires whoseinsulation is an increasing percentage of thewire area. With more and more turns of finerwire, the total area of the winding must in-crease if the desired copper area is maintained.When the maximum available window area isreached, improvement may still be obtained bygoing to more turns of finer wire, even thoughthe dc resistance will increase, if the reductionin ac resistance is sufficient. Otherwise, theonly solutions are: (a) Let the transformer runhotter, or (b ) use a larger size core which willprovide a bigger window ( and fewer turns are

usually required).

It is not unusual to require hundreds of fmewires in parallel to achieve the required dc andac resistances at 100 kHz or higher. Thisrequires that strands be twisted into largerstrands, and these twisted with each other ,progressing to a cable containing the totalnumber of fme wires needed to carry thedesired high frequency current. The effectivediameter of each strand is the circle of rotationof the outer extremities of the wires. Each levelof twisting further reduces the utilization factor .As shown in Table II, much better utilization isachieved when more wires/strands are twistedtogether at one time, because fewer levels oftwisting are required to achieve a similarnumber of wires.

For example, with 4 strands/twist, 4 wirestwisted comprise a Level 1 strand, four Level1 strands are twisted to obtain a Level 2 strandhaving 16 wires. Four Level 2 strands are thentwisted resulting in a Level 3 strand with 64wires, etc. Add levels until the desired numberof wires is reached.

The utilization factors of Table II are furtherreduced by pi/4 (.785) because round cablemade according to Table II occupies a squareportion of the winding space.

Instead of making up one cable containingall of the wires needed, it is often advantageousnot to twist the fmal level. This may providegreater flexibility in fitting the winding to the

TABLE II

Utilization Factor vs. Twist Levels

# Strands/Twist: l

~~

level 1 3 .65

level 2 9 .42

level3 27 .28

level 4 81 .18

level 5 243 .12

-L

~Y.ti!.:.

6 .67

36 .44

216 .30

1296 .20

..!L

~!llli..

4 .69

16 .47

64 .32

256 .22

1024 .15

i

~Y..ti!.:.

5 .69

25 .47

125 .32

625 .22

R2-12

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