LECTURE 33 Indutor Design 1 A. Overview of Copper versus Core Loss in Indutors 1. Core Material Limitations 2. Core Materials Compared 3. Filter Indutor Design via Erikson s Four Step Design Rules 4. Ten Commandments For Indutor Design 5. Summary B. Inlusion of Core Losses and Relation to Wire Winding Losses 1. Sinusoidal Flux Density B(wt) Driving the Core 2. B peak ~ vdt λ = NA (ore) NA THIS ALSO SAYS N~ 1/B that is as B dereases the number of turns will inrease 3. Tailoring B opt for Minimizing Total Losses: a. Trading Fe vs Cu to Ahieve Minimum Total Loss
2 b. N makes for larger I 2 R(wire) losses. N makes for higher B peak and more ore losses The various ore geometry s are shown below For a term paper on Integrated Indutor Design See as a starting point the artile LAYOUT OF INTEGRATED RF SPIRAL INDUCTOR in Ciruits and Devies Marh 1998 pgs 9-12
LECTURE 33 Indutor Design Methodology 3 A. Overview of Copper versus Core Losses For HW#1 do problems 14.1(easy) and 14.5(harder) An indutor is a devie whose purpose is to store and release energy. A filter indutor uses this apability to smooth the urrent through it and a two-turn flybak indutor employs this energy storage in the flybak onverter in-between the pulsed urrent inputs. The high µ ore allows us to ahieve a large value of L = µn 2 A /l with small A and l so large L values are ahieved in small volumes. However, high µ will limit the maximum energy storage in the ore with no air gap. Sine the magneti ore material itself is inapable of storing signifiant energy, energy storage is aomplished in a non-magneti air gap(s) in series with the ore. These gaps minimize the indutor variations aused by hanges in ore properties and help avoid ore saturation. If non-linear L(i) is desired, as it is in magneti amplifers, it an also be ahieved with a stepped or tapered air gap as shown below: φ Air Gap Both ore and wire winding losses as well as saturation effets establish design rules for those who wind their own indutors as shown below. Limitations of magneti ores are ruial to good indutor design 1. Core Material Limitations: In d appliations, indutors are primarily thought of as urrent operated devies. Too large a urrent, will saturate the magneti ore via ampere-turn limits set by ni< B SAT A C R, this means ni <B SAT /µ. This limit on I MAX will in turn limit the maximum energy that an be stored in an ungappped ore to
B 2 (SAT) l(ore)a(ore) /2µ. A onvoluted inter-dependene of the wire urrent and the magneti field in a transformer ore is as follows. The volt-se limit says that B(ore)= λ p = V L dt /N A C. Hene, we find that N~1/B and later we will employ N 2 ~ 1/B 2 in relations for the opper loss to show P Cu ~1/B 2. As regards the indutor voltage applied without onsidering the equivalent series resistane r L, i L on a DC basis will ontinue to inrease with a DC V L,sine all of the applied voltage appears aross v L. With r L the indutor urrent inreases until i L r L equals the input voltage. Again the B-H urve of the indutor ore looks like a E L dt versus I L urve as shown below. At high Edt,I where the ore saturates and L 0. For a flux or B hange to our a time interval is required over whih the applied indutor voltage is applied. 4 In high frequeny PWM appliations, the major magneti ore material limitations are: 1. Flux Density Saturation and 2. Core losses, both of whih depend upon flux swing and applied frequeny of the flux. In these appliations indutor windings are usually driven with retangular voltage waveforms derived from low impedane soures. Sine the voltage, pulse width, and number of turns are quite aurately known, it is easy to apply Faraday s Law to determine the maximum flux swing and appropriately limit it.. In the ase of a sinusoidal voltage, Faraday s law gives V= N dϕ / dt =Nω A C B SAT. Hene V/ωNA must be less than B SAT.
A flybak transformer is atually an indutor with multiple windings. It stores energy taken from the input in its mutual indutane during one portion of the swithing period, then delivers energy to the output during a subsequent interval. In a flybak transformer, the magnetizing urrent is virtually important, beause it represents the energy storage required by the appliation. In this ase, the magnetizing urrent an be alulated quite aurately using Ampere s Law, beause it depends on the very preditable harateristis of the gap in series with the ore, and the unertain ore ontribution to energy storage is negligible. Magnetizing indutane is an essential element in a flybak transformer. 2. Core Materials Compared There are three major ore materials of interest. Metal Alloy Tape Powdered Metal Ferrites 1. B sat = 1 Tesla 1. B sat = ½ Tesla 1. B sat = 0.2 Tesla 2. µ r = 60,000 2. µ r = 100 2. µ r = 1000 3. ρ is low, high eddy 3. Highest Loss Core 3. ρ is high, low eddy urrent losses 4. Sharp Saturation suddenly L 0 4. Soft Saturation L(1/4 tesla) = ½L(B=0) Hene I m inreaes gradually urrent losses 4. Sharp Saturation suddenly L 0 For a slotted ore indutor the winding your own L magneti ore looks like that below. Most energy storage ours in the air gap region as we showed in the prior leture. 5 W(air) = ½B 2 l g /A g µ o >> W(ore) = ½B 2 l /A µ
The above air gap allows lots of leakage flux to impinge on the wire windings. For a filter indutor this is not an issue. But for indutors with very high AC urrent omponents this leakage flux an ause very substantial inreases in the AC wire resistane, via previously disussed proximity effets. We an redue the level of leakage flux by a different ore design as show below. The air gap in an E type ore is inside the ore enlosure and if the g/d ratio is hosen small, the short enter leg providing the air gap will produe little fringe flux as shown below: 6 g d d 2d 4d d I DC >> I a Case for Filter Indutors I ritial I DC > I a r L L t I L P u (loss) > P ore (loss) sine B AC is negligable 2 I (rms) r > ore loss L L I a >> I DC Case for Ciruit Indutors in PWM Converter Ciruits: I a << I(rititial)
7 r L L I a dominates t P total loss = P u (loss) + P ore (loss) Herein we will learn to trade Fe vs Cu in order to minimize the total loss. That is we an find an optimum number of turns of wire that provide minimum total loss. If N is too high I 2 R wire losses dominate. While if N is too low B peak = Edt/NA will rise and ore losses dominate. The question is now how do we rationally trade off opper losses and ore losses to ahieve the desired indutor for the speifi iruit use. While the graph below illustrates the balane needed to ahieve minimum total loss, we will not fully understand this plot till letures 34 35. P T More Cu loss P u ~ 1/B 2 Surprise! result only true for transformers B opt for minumum loss B As B we find More Core losses P ore ~ B 2 or B x as expeted 3. Filter Indutor Design via Eriksons Four Step Design Rules We tailor the indutor design to the swithed mode onverter iruit topology. Design onsiderations for indutors inlude: ore size and permeability effets of air gaps on relutane and L linearity as well as the sensitivity of L to hanges in the ore ore loss limits saturation values of flux density in the hosen material ore window size (for allowing all needed wire windings) urrent density allowed in wire windings
maximum values of a voltage, d volt-se, and d urrent operating frequeny 1 st Selet ore via the emperial K g fator Both hysteresis and eddy urrent ore loss effets in indutors are aused by time-varying flux. If an indutor arries a onstant d urrent below its saturation limit, the ore flux will be onstant, and the hysteresis and eddy urrent losses of the ore will be zero. In pratie any swith mode onverter has large urrent ripple in the indutors, at the swithing frequeny. The flux variation will be follow the AC urrent. Magneti material an handle only a limited loss per unit volume without getting too hot. Losses of 1 W/m 3 are usually onsidered high for ferrites, and values in the range of 0.2 W/m 3 are more ommon in well designed ores. At any rate we an translate iruit variations into a single parameter whih helps to speify the ore needed as shown in Leture 32 2 2 max 8 5 K g > L I ρ(wire) * 2 10 [ m ] B R K pk u 8 Get L, R, I max from PWM onverter eletrial spes Get B pk, K u from manufaturers ore and wire spes respetively. The ore geometry gives us both A (ore) and W A (wire winding area) as well as the mean length of turn wound on the ore. 2 nd Spe the air gap to be ut in the indutor ore lg µ 2 o L I max = B A [m] pk As an alternative to the gap size, l g, we an speify the A L, or speifi indutane in mh per turn 2, fator for an indutor ore. N w is the number of wire turns and R (ore) is ore relutane inluding the airgap. A L L N 2 w 1 = R 1 = R mh turn - We arbitrarily fix N @ 1000 turns as ore manufaturers do so in their A L speifiations. 1 < A L < 1000; A L is typially 100 for N = 1000 turns.
You spe A L - Core Manufaturer will ut l g to size to meet your spe s 9 3 rd Spe # of wire turns. No frational turns are allowed making this an iterative proedure N = L I max w B A pk 4 th Choose wire insulation, wire type and urrent arrying area or AWG# size K u(window area) Area Wire < N All the wire wound on a given ore must fit into its winding window, the opening for wire turns. The window also must hold insulation and any struture on whih the wire is mounted. In pratie, only about 50% of the window area an atually arry ative ondutor, as the rest is voids or wire insulation. This fration is alled fill fator. In a two-winding transformer, this means that eah winding an fill not more than 25% of the total window area. That is given the wire winding area, W A, and the required number of turns we set the required wire size. We usually use AWG # s instead of wire urrent area to speify a wire. This hoie of wire size will be modified by high frequeny skin effets as we will show later. Wire size is an important aspet of the indutor design sine a given wire an handle only a limited urrent density to avoid exessive power loss. The wire-winding window of a given ore must have enough area so that opper wire of a given diameter an be used and all the required number of turns fit. We do this to avoid exessive Ohmi heating of the wire so wire loss influenes key details in magneti ore geometry hoies. For an indutor design, ore saturation primarily limits the amp-turn values possible but the urrent density limit in the wire also represents an ampturn limit. A trade-off rule results. Consider what happens if the rule is violated. If the ore window is too small, the wire size hosen ould be too small for the required urrent
and the wire will overheat before the saturation amp-tum limit of the magneti ore is reahed. If the wire-winding window is too large, then ore saturation will be reahed prematurely and the opper winding might be underutilized. Core size sets the mean length of turn required to enirle the ore and hene the length per turn of the wire. R wire = (resistane per meter of the wire) x (length per turn) X N Or alternatively R=ρN (MLT)/A W (wire). MLT depends on ore geometry. Cores made of setioned strutures tend to be easier to wind with automated wire winding mahines than toroids. For toroid ores, windings are often designed to form a single layer of opper material around the inside of the ore. This keeps the devie small and minimizes flux leakage. For other ore shapes, windings often use the largest wire that will fit onveniently into the window. This minimizes losses, and maximizes the power rating. 3. General Indutor Design via Ten Design Rules(SKIP THIS SECTION IF YOU ARE AN UNDERGRADUATE) a. Overview of ten step L design sequene 10 Start L I rms and f required by PWM onverter iruit. 1. T(ore surfae) }In steps 1-2 the }Energy and heat 2. Stored energy required}flow requirements ~1/2 Li 2 }help hooses both }material type and 3. From tables of ores vs}geometri shape 1/2 Li 2 }of the required }ore to wind wire 4. Thermal Analysis }upon. T(surfae), T(ambient)
Steps 5 and 6 quantify: B a waveform in the ore from ore data speifies eddy urrent ore loss 11 Peak B is ruial to hysteresis loss B(peak) < B sat Winding parameters for opper wire Maximum L max = N 2 /R(ore) }L desired < L max }L=N 2 /(R(ore)+ R(gap)) }8-10 set }the L }value by }µ(ore) }and }l g (gap) b. Detailed Approah for Eah of 10 Steps For HW#1 verify the speifi quantities in the example below on the left hand side. 1 fl Assemble Indutor design inputs STEP 1 Six Design Inputs Speifi (1) L values desired by iruit: (1) L = 300 µh with 4A rms (2-3) I rms and I peak : I peak = found from
depends on the 2 I rms only for a sine wave. Current waveshape eg. I p = 2 4 = 5.6 (4) Swith Frequeny We hoose 100 KHz operation (5) Max T of ore surfae T s = 100 o C (6) Max ambienttemperature T a = 40 o C The above Six parameters enter into the indutor design inputs of step one. Key is the design produt: L I $ I stored energy rms p. We get both I rms and I p (peak) from I L vs time waveforms. This sets the both required ore material and geometri ore shape required to dissipate the heat as shown below. General Instead of STEP 2 fl Compute L $ I Irmsin units of H-A 2 Speifi ε L = 1 2 L 2 I rms LI rms I peak for a sinusoid with 12 Use LI rms I max Better aounts for odd waveforms found in PWM onverter iruits the hosen Lis.007 H-A 2 for this ase µ and B SAT alone doesn t fully speify the ore as we must also onsider shape and size of the ore to aommodate windings and heat flow ore material - Choie of swithing f is also ruial
- to math to the ore ore size - B max < B sat ore shape - ease of putting N windings of wire - on the iron (ore ost as well as wire ost) is also an issue in some low ost ommodity iruits. STEP 3 fl Choose ore material, shape and size 13 We need a manufaturers ore data base to guide our possible ore hoies. Below we list just the pertinent data for this 100 khz example Nφ = Li = flux linkage Where N w = K Awindow RMS u J rms = I A u Au φ = B A ore A opper = A u K Cu an vary from 0.3(Litz wire) to 0.9 for foil. We will need a larger ore for the hoie of Litz wire ands a smaller ore for foil wire. Wire type with a opper fill fator will need to be balaned with ore size. Now we reonile eletrial iruit spes and the wire and ore speifiations. Ciruit Spe s Core and Winding Speifiations LI rms I(peak) Five Parameters K u, J rms, B pk, A w, A Design spes Two Materials Issues: J rms, B pk
from iruit Three Geometry Choies: A w, A, K u K u is derived from wire size,the number of turns and the ore wire winding area and is not an independent variable. It varies by a fator of 3 from Litz wire K u 0.3 to Foil K u 0.9 STEP 4 Thermal Resistane, R θ, and Power loss Assume the total indutor power: P L P ore + P winding. As a first guess assume that that ore and wiring ontribute equal parts. Ts - Ta Realize that P L (max) is limited via thermal heat: R(ore total) We annot allow the ore to reah 100 C from the balane of heat versus heat taken away. LI rms I peak ~ K u [J rms B pk A w,a ] Note that the type of wire must be hosen at his point but not wire size. A speifi hoie is Litz wire with K u = 0.3. Note the trends set in motion by K u hoie. K u then required ore size then required ore size This is lassi Trading ore for opper in the indutor design. 14 General Speifi P Ts - Ta R θ (ore) From ore data base or by alulating R θ from geometry P = P m V T Typially for a ore V = Core windings plus Core Volume, R θ ~ 10 o /W V T = V + V w In General Core loss P m varies with the flux density to some power and P m inreases with frequeny as shown below.
P m (loss) mw/m 3 f 15 B pk (T) From P M vs B pk vs f or from the hosen ore data base we find B and then P m. Where P m is Speifi Power density. Take as illustrative for 3F3 ore under the above onditions 250 mw/m 3. Next we do two steps together: speifing B(ore) and B peak 5 fl Speify ore flux density B a 6 fl Find allowable maximum ore flux density β $ This depends on our hoie for ore material and upon the ore temperature. For fixed (T s -T A ) and given frequeny of applied urrent we find from the loss equation P M ~ k (B a ) 2 / f 2 We find B a by employing a maximum ore temperature of 100 degrees and knowing the ore thermal impedane. The allowed B a (max)= 170 mt sinusoidal for the ase we are disussing here. B(wt) = B a Sinwt B a,while 1.414xB a B pk. Here we have to insure that B max does not exeed B SAT for the hosen ore material. Otherwise we have to iterate with a new ore or new wire or both as shown below in step 7. The flux density in the ore is proportional to the indutor urrent. STEP 7 fl Design wire winding (k u, J,
A u, N) 16 General Speifi Au 1. K u = N 1. Eah type of wire has Awindow We use W A for ore wire window and K u is set from hoie of wire type. We previously hose 0.3 Litz wire K u = 0.3 for Litz Wire This allows us to set W A for the ore when the number of wire turns is known Irms 2. A u from wire database 2. A u = 4 A Jrms Jrms I rms omes from iruit waveforms J rms from wire data sheet J rms omes from wire data base and 4A from iruit spe. We utilize: K Cu =0.3, and find first J RMS and then the required wire area at the given urrent level, 4A. A u as determined form 4A/ A u = 6A/mm 2 This is shown on the next page.
General Speifi The wire winding losses in Watts per m 3 is next sought 3. P M = P windings = 22 k u J 2 rms mw m 3 3. rms J = ku 6A/mm 2 Given P m we find J rms. This sets the wire area, A Cu =4A/6A/mm 2 =0.7mm 2 We also know from the ore data A(window) =140 mm 2 so we know N = K u Awindow N =.3140 = 60 + Au 0.7 Note the importane of A window = 140 mm 2 from ore data base for the number of turns of opper wire. General L max N A B I peak pk 3.3 Required # of opper wire turns results! 8 fl Find maximum indutane of seleted magneti ore Speifi A =1.5 * 10-4 m 2 from hosen magneti ore data base = N, B pk and I p are all given + -4 L max must > L spe to insure low i ripple L max = 60 1.5*10 170 4 2 But not too muh bigger beause = 290 µh bigger ore is more ostly. 17
General L ~ N 2 R, lore R ~ µ Aore If L max < L spe then Choose a bigger ore R But if L max >> L spe we have hosen too big a ore. Save $ by reduing ore size. Speifi 18 L spe is 300µH so we are under target. What to do? So lose just use no air gap with ore or hoose smaller ore size with tailored gap. 9 fl Design airgap length g Set the ore air gap g so B pk < B sat (for hosen ore) B sat varies only 0.1 2 Tesla as we hoose various ore materials. If L max > L spe then we an tailor L spe from L max by adding preision air gaps. The gap relutane negleting fringing is: lg Rgap = µ o Ag l g usually < 1 mm Preision mylar sheet @ 1/2 mil is available when we lamp two piees together to form a gap. Notebook paper is 3 mil / 3 thou
Double E ore has three distributed air gaps -3 air gaps are better than one Why? Pbm of one big gap is that B(fringe) from the ore goes out further into opper windings inreasing skin effet and eddy urrent loss in the wires. 19 The effetive air gap with an E ore geometry is made in a ross-setion and negleting fringing. Σ gaps = 3 * E ore spaing The area of the gap A g = Make 1/3 of size for less fringing A g (a+g)(d+g) gap size g << d,a - ore sizes if flux extends out additional g/2 from ore edge A g ad + g(a+d) + 0 seond order terms R total = R (ore) + R gaps l 3g µ A µ o Ag Usually assume R is small ompared to R(air gap). We tailor 3 g to give desired B pk (ore) when I pk is applied so that we do not exeed B(ritial). NI = R g φ NIpk 3g = A B µ A pk o g Tune g for desired B pk so B pk < B sat (ore)
10 fl Set L to design value 20 if L spe < L max We have two routes to redue L max to reah L spe. L = N2 Rg We ould inrease R g alone but this both redues B pk and B a whih is good only for ore loss. Sine R g = 3g by reduing A g we ould use µ o Ag a smaller ore, thereby reduing L and lowering ore ost. If we redue N 2 to reah L spe from L max, then we save opper. But lower N inreases B pk and hene ore loss again we trade: Cu for iron.
Summary Start 21 B. Calulating Magneti Core Loss 1. For Assumed Sinusoidal B(wt) Exitation The total power loss inluding both hysteresis and eddy urrents.
P W per x y m = 3 k fe (B pk) f Vore m. Where V ore = A l The exponent x depends on how wide the hysteresis loop expands horizontally (AH) as well as vertially (AB). 1.5 < X < 3.0; 2 is typial of pratial ore materials 1 < Y < 2; depending on ore material 22 0 < P m < 250 mw m 3 B max is the key parameter P T = P m * V ore = P m A l Most PWM onverter i L waveforms are square or triangle waves not sinusoids. These signals ontain DC omponents. B(d) auses no ore losses only ore saturation! Harmonis of the swith frequeny also our for many onverter waveforms Higher ore losses due to the f 2 dependene of eddy urrent losses. 2. Restraining B peak Values to below B(ritial) We now onsider voltages applied aross the indutor, rather than urrents driven through the indutor. Volt-se balane will be onstrained
by B SAT of the hosen ore material as shown earlier. a. A soft B-H urve B With soft B-H urves L will derease slowly with urrent and B sat not hange preipitously. Often L(B sat /2) = L((B=0)/2) for soft saturation ores I max H Ni 23 However above B sat µ µ o and L short iruit or very small values Then for B>B SAT we find L = µ o A N 2 where we replae µ(ore) by l µ(air) and L is 1000 times smaller or so b. B-H is also v L dt vs i L I V in magneti ore N wire turns V L = L di L dt VL dt i L = L N i L = φ R, φ = B A For V in to the indutor being a sinusoid or any waveform the φ inreases to a peak only after integrating V L. A step of V L is easiest to visualize. v L i φ = Ni = N v dt L L 2 R N µ l t A Where R = µ l A N 2
The flux varies as v Ldt µ N A l v dt φ L = BA. Now B pk ours when i L and φ peaks. N vdt λ 1 B pk = = (volt-seonds) N A N A The maximum B pk ours for when V L is either +V L for a long period of time and the indutor urrent is ramping up to high values. The V L and +V L square wave exitation vs. time auses a triangular B pk waveform vs time. t +B pk 24 v L λ( v se) B max 2 N A This integral dependene of B on V L says: i L 2B max 1. B max ~ 1 This relates the hoie of wire turns to ore issues. N It says for many wire turns B max while for one wire turn we ahieve maximum B whih must not exeed B SAT. Hene, it sets a minimum number of wire turns required. 1 2. B max ~ This is sets up the need for larger size ores. A There are relations between the opper loss and ore loss. 3. N B max redues iron loss } Cu-Fe loss trading ours N inreases I 2 R Cu loss } trade again! We then have to distinguish the ases of urrent through an indutor, amp-turn limits, and the ase of voltage impressed aross an indutor, -B pk
termed volt-se limits. Eah sets limits to indutor performane and in some ases they at in onert to limit indutor operation. 25 Summary of Saturation Design Rules: Rule Interpretation Ni < B sat RA Amp-turn limit for an indutor W max = ½B 2 sat l ore A ore /µ Maximum energy that an be stored in a given ore. W max = ½Bsat 2 V gap /µ o Maximum energy is determined by air gap volume if the ore has high µ. V o /N < wb sat A Maximum volts per turn (for a transformer) at frequeny w. vdt < NB sat A Maximum volt-seonds for an indutor or transformer.