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Section 5

Filter , boost inductors and flyback transfonners are all members of the "power " family. They all function by taking energy from the electrical circuit, storing it in a , and subsequently returning this energy (minus losses) to the circuit. A flyback transfonner is actually a multi- winding coupled inductor, unlike the true transfonn- ers discussed in Section 4, wherein is undesirable.

Application Considerations Design considerations for this family of inductors vary widely depending on the type of circuit applica- tion and such factors as operating frequency and rip- ple current. Inductor applications in switching power supplies can be defined as follows (see Fig. 5-1):

.Single winding inductors: Output filter inductor (buck-derived) Boost inductor Flyback (buck-boost) inductor Input filter inductor .Multiple winding inductors: Coupled output filter inductor (R5) Inductor design also depends greatly on the in- ductor current operating mode (Figure 5-2): Design limitations: The most important limiting .Discontinuous inductor current mode. when the factors in inductor design are (a) temperature rise and instantaneous ampere-turns (totaled in all wind- efficiency considerations arising from core losses and ings) dwell at zero for a portion of each switching ac and dc winding losses, and (b) core . period. Output filter inductors (buck-derived) --single .Continuous inductor current mode. in which the and multiple windings are seldom operated in the total ampere-turns do not dwell at zero (although discontinuous current mode because of the added the current may pass through zero). burden this places on the output filter , and because it results in poor cross-regulation in multiple In the continuous current mode, the ripple current output supplies. Typically operated in the continuous is often small enough that ac winding loss and ac core mode with peak-peak ripple current much smaller loss may not be significant, but in the discontinuous than full load current, ac winding loss is usually not mode, ac losses may dominate. significant compared to dc loss.

5-1 design is then usually limited by dc winding losses and core saturation. However, many boost and flyback applications are designed to operate in the discontinuous mode, because the required value is less and the inductor physical size may be smaller. But in the dis- continuous mode, the inductor current must dwell at zero (by definition) during a portion of each - ing period. Therefore, the peak of the triangular cur- rent waveform, and thus the peak-to-peak ripple must be at least twice the average current, as shown in Fig. 5-2(a). This very large ripple current results in a po- tentially serious ac winding loss problem. Also, the resulting large flux swing incurs high core loss. Core loss then becomes the limiting factor in core utiliza- tion, rather than saturation, and may dictate a larger core size than otherwise expected. Thus, the circuit designer's choice of operating mode makes a substantial difference in the inductor design approach. When flyback transformers are operated in the continuous inductor current mode, the total ampere- turns of all the windings never dwell at zero (by defi- nition). However, the current in each winding of any flyback transformer is always highly discontinuous, For example, assume full load Idc of 10A, and regardless of inductor current mode. This is because typical peak-peak triangular ripple current 20% of Idc, current ( ampere-turns ) transfers back and forth be- or 2A (worst at high Vin). In this example, the worst- tween primary and secondary(s) at the switching fre- case rms ripple current is 0.58A (triangular wave- quency. As shown in Fig. 5-3, the current in each winding alternates from zero to a high peak value, form rms equals I pp / .Jj2 ), and rms ripple current even though the total ampere-turns are continuous squared is only .333, compared with the dc current with small ripple. This results in large ac winding squared of 100. Thus, for the ac f R loss to equal the loss, regardless of the operating mode. dc loss, the RajRdc ratio would have to be as large as However, the core sees the total ampere-turn 300 (Section 3, Fig. 3-5). This is easily avoided. ripple. Thus, core loss behaves in the same manner as Therefore, ac winding loss is usually not significant. with the single winding flyback inductor discussed Also, the small flux swing associated with small previously- small core loss when designed and oper- ripple current results in small core loss, with high ated in the continuous mode, large core loss in the frequency core material operating below discontinuous mode. 250kHz. Core utilization is then limited by saturation (at peak short-circuit current). However, the small flux swing may permit the use of lossier core materi- als with higher BSAT,such as powdered , Kool- mu@, or laminated metal. This may enable reduced cost or size, but core loss then becomes more signifi- cant. Also, distributed-gap materials exhibit rounding of the B-H characteristics (Sec. 2, pg. 2-3), resulting in decreasing inductance value as current increases. Boost and input filter inductors and single winding flyback inductors are often designed to operate in the continuous mode. As with the buck- derived filter inductors described previously, inductor

5-2 Losses and Temperature Rise dered metal cores, the winding(s) should be likewise distributed. Thus, a toroidal core shape should have The discussion in Section 4 regarding tempera- the windings distributed uniformly around the entire ture rise limits, losses and thermal resistance in trans- core. formers (pp 4-1,2) is generally applicable to induc- With a discrete gap, used with laminated metal tors, as well. alloy cores or ferrite cores, the winding should be Balancing Core and Winding Losses directly over the gap. For example, if a pair of "C" When inductors are designed for the discontinu- core halves has a gap in one leg and the winding is ous mode, with significant core loss, total loss is at a placed on the opposite (ungapped) leg, as shown in broad minimum when core and winding losses are Fig. 5-4a, the entire magnetic force introduced by the approximately equal. But when inductors are de- winding appears across the two core halves. This re- signed for the continuo~s mode, core loss is often sults in considerable stray flux propagated external to negligible, so that the t~talloss limit can be allocated the device, in addition to the flux through the gap. entirely to the windings. The energy stored in the external stray field can eas- General Considerations --Core ily equal the energy stored in the gap, resulting in an inductance value much greater than expected. The Ideal magnetic materials cannot store energy. external stored energy is difficult to calculate, making Practical magnetic materials store very little energy, the total inductance value unpredictable. Also, the most of which ends up as loss. In order to store and additional flux in the stray field will cause the in- return energy to the circuit efficiently and with mini- ductor to saturate prematurely. mal physical size, a small non-magnetic gap is re- However, when the same winding is placed on quired in series with a high permeability magnetic the gapped core leg, as in Fig. 5-4b, the entire mag- core material. In ferrite or laminated metal alloy netic force introduced by the winding is dropped cores, the required gap is physically discrete, but in across the gap directly beneath. The magnetic force powdered metal cores, the gap is distributed among across the two core halves is then nearly zero, and the metal particles. there is little external flux. The core then serves its Paradoxically, virtually all of the magnetic en- intended purpose of providing an easy (low reluc- ergy is stored the so-called "non-magnetic" gap(s). tance) return path for the flux, requiring very little The sole purpose of the high permeability core mate- magnetic force to do so, and propagating very little rial is to provide an easy, low reluctance flux path to external field. link the energy stored in the gap to the winding, thus efficiently the energy storage location (the gap) to the external circuit. In performing this critically important function, the material introduces problems: (a) core losses caused by the flux swings accompanying the storage land release of energy, and (b) core satu- ration, where the core material becomes non- magnetic and therefore high reluctance above a cer- tain flux density level. The energy storage capability of a practical gapped core is thus limited either by temperature rise associated with core loss, or by core saturation. Stray Flux. Another problem that must be faced is stray flux, associated with energy stored in a fringing field outside the gap. Stray flux couples noise and EMI to the external circuit and to the out- side world. This stray energy also increases the in- ductance b(!yond its intended value by an amount that is difficult to predict. To minimize stray flux, it is very important that the winding distribution conforms to the gap. When the gap is distributed throughout the core, as in pow-~

5-3 Gap area correction: Even when the winding is Melted windings: Another serious problem can properly placed directly over a discrete gap, there will result from the fringing field adjacent to the gap. Any be a small but intense fringing field adjacent to the winding turns positioned close to the gap will likely gap, extending outward beyond the boundaries of the exist within the high flux density of the fringing field. core cross-section as shown in Fig 5-4b. Because of In applications with large flux swings, huge eddy cur- this fringing field, the effective gap area is larger than rent losses can occur in those few turns close to the the core center-pole area. To avoid what could be a gap. Windings have been known to melt in this vicin- significant error, the inductance calculation must be ity. This problem is most severe with flyback trans- based upon the effective gap area rather than the ac- formers and boost inductors designed for the discon- tual center-pole area. An empirical approximation is tinuous mode, because the flux swings at full load are obtained by adding the length of the gap to the di- very large. With filter inductors, or any inductors de- mensions of the core center-pole cross-section.(l) signed for continuous mode operation, flux swing is For a core with a rectangular center-pole with much less and the problem is much less severe. cross-section dimensions a and b, the effective gap Solutions for devices designed to operate with area, Ag is approximately: large flux swing: (1) Don't put winding turns in the immediate vicinity of the gap. Although the winding Acp=axb; Ag~(a+J!g)x(b+J!g) (la) should be on the center-pole directly over the gap, a non-magnetic, non-conductive spacer could be used For a round center-pole with diameter Dcp. to substitute for the turns in the area where the fring- ing field is strong. (2) Distribute the gap by dividing it into two or three (or more) smaller gaps spaced uni- formly along the center-pole leg under the winding. Since the fringing field extends out from the core by a distance proportional to the gap, several small gaps will dramatically reduce the extent of the fringing field. This also results in more accurate inductance (lb) calculation. (3) Eliminate the fringing field entirely by using a with a powdered metal rod Thus, when £g equals O.lDcp, the area correction substituted for the ferrite center-Ieg. This distributes factor is 1.21. The gap must be made larger by this the gap uniformly among the metal particles, directly same factor to achieve the desired inductance (see Eq. beneath the entire length of the winding, eliminating 3a). the fringing field. While this last method has been The preceding correction factor is helpful when used successfully, it is usually not practical because the correction is less than 20%. A more accurate cor- of high cost, and greater ac core losses with metal rection requires finite element analysis evaluation, or powder cores. trial-and-error evaluation. Gapping all legs: It is tempting to avoid the cost After the number of turns and the gap length have of grinding the gap in the centerleg by merely spacing been calculated according to steps 7 and 8 of the the two core halves apart, thus placing half the gap in cookbook design procedure presented later in this the centerleg and the other half in the combined outer section, verification is obtained by building a proto- legs. But the outer leg gaps clearly violate the princi- type inductor. ple that the winding should be placed directly over If the measured inductance value is too large, do the gap. A little more than half of the total magnetic not reduce the number of turns, or excessive core loss force will exist across the centerleg gap, but the re- and/or saturation may result. Instead, increase the gap maining force appears across the outer leg gap(s) and to reduce the inductance. thus across the two core halves. This propagates con- If the measured inductance is too small, the num- siderable stray flux outside the inductor, radiating ber of turns may be increased, but the core will then EMI, and the inductance value becomes larger and be under-utilized and winding losses may be exces- difficult to predict. The result is intermediate between sive. It is best to raise the inductance by decreasing Figures 5-4a and 5-4b. the gap length.

5-4 There is one other benefit of spacing the core As discussed in Section 4, pot cores and PQ cores halves apart. Because the gap is divided, the smaller have small window area in relation to core size, and centerleg gap length reduces the fringing field and the window shape is not well suited for flyback trans- thus reduces the problem in the winding fonners or discontinuous mode inductors. close to the gap. EC, ETD, LP cores are all E-E core shapes, with A trick which greatly reduces the external stray large, wide windows which make them excellent flux in this situation is to place an external shorted choices with ferrite materials. These core shapes lend turn around the entire outer periphery of the inductor. themselves to spiraled windings of wide copper strip, The shorted turn is made of wide copper strip placed especially with inductors operated in the continuous co-axial with the inductor winding, encircling the mode, where ac winding losses are small. entire outer surface of the inductor, outside the Toroidal powdered metal cores, with windings windings and outside the outer core legs, and closely distributed unifonnly around the entire core, can be conforming to the external shape. Any stray flux that used in any inductor or flyback transfonner applica- escapes to the outside world will link to this external tio~. Stray and EMI propagation is very shorted turn, inducing in it a current which creates a low. magnetic field in opposition to the stray flux. But a gapped ferrite toroidal core is a very bad choice. Windings distributed around the toroid will Core Selection: Material not confonn to discrete gaps, resulting in large stray Select a core material appropriate for the desired fields, radiated EMI, and inductance values that can- frequency and inductor current mode. not be calculated. Ferrite is usually the best choice for inductors de- signed to operate in the discontinuous mode at fre- Optimum core utilization quencies above SOkHz, when core loss associated The smallest size and lowest cost inductor is with large flux swing limits core utilization. achieved by fully utilizing the core. In a specific ap- However, in the continuous mode, with small plication, optimum core utilization is associated with ripple current and small flux swing, ferrite cores will a specific optimum gap length (resulting in a specific often be limited by saturation. In this case, lossier effective penneability fie for cores with distributed core materials with greater saturation flux density, gaps). The same core in a different application or at a such as powdered iron, Kool-mu@, powder, different frequency may have a different optimum or even gapped laminated metal cores may enable gap length. reduced cost or size. But the rounded B-H character- The optimum gap length results in the core oper- istic of powdered metal cores can result in the per- ating at maximum flux density (limited either by satu- haps unintended characteristic of a "swinging choke," ration or by core loss), and also at maximum current whose inductance decreases at higher current levels. density in the windings (limited by winding loss). This is the best possible utilization of the core, re- Core Selection: Shape sulting in the smallest size. The inductor design ap- The core shape and window configuration is not proach should therefore seek to achieve this optimum critically important for inductors designed to operate gap length (or optimum fie for a core with distrib- in the continuous mode, because ac winding loss is uted gap). usually very small. Figure 5-5 shows the characteristic of a core with But for inductors designed for discontinuous optimum gap, limited by core saturation and by max. mode operation, and especially for flyback trans- in the windings. The area between the formers, the window configuration is extremely im- characteristic and the vertical axis indicates the en- portant. The window should be as wide as possible to ergy storage capability. Any other slope (different maximize winding breadth and minimize the number gap size) results in less energy storage capability of layers. This minimizes ac winding resistance. For a flyback transformer, the wide window also minimizes Core Selection: Size inductance, and the required creepage dis- The discussion of core size for transfonners in tance when line isolation is required has less impact. Section 4-8 is mostly relevant to inductors as well. With a wider window, less winding height is re- The following Area Product fonnulae are intended to quired, and the window area utilization is usually help provide a rough initial estimate of core size for better. inductor applications.

5-5 tor, KpRI is the ratio of the total copper area to the window area, Aw. For a flyback transfonner, KpRI is the ratio of the primary winding copper cross-section area to the total window area. The saturation-limited fonnula assumes winding losses are much more significant than core losses. Kl is based on the windings operating at a current den- sity of 420A/cm2, a commonly used "rule of thumb" for natural convection cooling. In the core loss limited fonnula, core and winding losses are assumed to be approximately equal. Ther~- fore, the winding losses are halved by reducing the current density to 297A/cm2 (470 x 0.707). Thus, K2 equals 0.707.Kl. In either fonnula, it is assumed that appropriate Fig 5-5 Optimum Core Utilization techniques are used to limit the increase in winding losses due to high frequency to less than When core loss is not severe, so that flux swing is 1/3 of the total winding losses. limited by core saturation: Forced air cooling permits higher losses (but with reduced efficiency). K values become larger, result- 41 ing in a smaller core area product. (2a) The 4/3 power shown in both area product for- mulae accounts for the fact that as core size increases, the volume of the core and windings (where losses are generated) increases more then the surface area (where losses are dissipated). Thus, larger cores must (2b) be operated at lower power densities. For the core loss limited case, L\BMAXmay be ap- where: proximated by assuming a core loss of lOO mw/cm3 - L = inductance, Henrys a typical maximum for natural convection cooling. Iscpk = max pk short-circuit current, A For the core material used, enter the core loss curves BMAX = saturation limited flux density, T at 100 mw/cm3 (Fig. 2-3). Go across to the appropri- M = current swing, Amps (primary) ate switching (ripple) frequency curve, then down to L\Bmax= max flux density swing, Tesla the "Flux Density" scale (actually peak flux density). IFL = fillS current, full load (primary) Double this number to obtain peak-peak flux density, K1, K2 = JMAXKpRI x 10-4 L\BMAX. If units are in Gauss, divide by 10,000 to convert L\BMAXto Tesla, then enter this value into the where: JMAX = max. current density, Alcm2 core loss limited Area Product fonnula. KpRI = primary copper area/window area In filter inductor applications, nonnally operated 10-4 = converts dimensions from meters in the continuous inductor current mode, the ripple tocm current is usually only 10-20% of the full load dc cur- rent. Ferrite cores will usually be limited by satura- Application KpRI KI K2 tion flux density, not by core loss, at switching fre- Inductor,single winding 0.7 .03 .021 quencies below 250 kHz. Boost and flyback induc- FilterInductor, multiple winding .65 .027 .019 tors, and flyback transfonners operated in the con- Flybacktransformer -non-isolated 0.3 .013 .009 tinuous current mode, where total ripple ampere-turns Flybacktransformer -with isolation 0.2 .0085 .006 are a small fraction of full-load ampere-tums, may also be saturation limited. In these situations, it may For a single winding inductor, the tenn "primary" be possible to reduce size, weight, and/or cost by us- above refers to the entire winding. ing core materials that are lossier but have higher KpRI represents the utilization of the window saturation flux density, such as Kool-Mu@, or metal containing the winding. For a single winding induc- alloy laminated cores.

5-6 However, when these applications are designed Inductance Factor, AL, expressed in milliHen- for discontinuous mode operation at full load, ripple rys/1OOOturnS2, or nanoHenrys/turn2, is often stated ampere-turns are so large that the inductors will al- by the manufacturer for pre-gapped ferrite cores or most certainly be core loss limited. for distributed-gap powdered metal cores. It provides If uncertain whether the application is core loss a convenient method for calculating inductance for an limited or saturation limited, evaluate both formulae existing gapped core with a given number of turns, and use the one which results in the largest area but it is awkward for determining the optimum gap product. length or the optimum effective permeability for best These initial estimates of core size are not very core utilization. accurate, but they do reduce the number of trial solu- tions that might otherwise be required. The detailed L = N2 AL nanoHenrys (3c) design process provides greater accuracy. In the final In the inductor design process, the desired in- analysis, the validity of the design should be checked with a prototype operated in the circuit and in the ductance is presumed to be a known circuit value. environment of the application, with the hot spot The optimum gap length, Rg, or effective permeabil- temperature rise measured by means of a thermocou- ity , jJe to achieve that inductance is calculated by in- pie cemented alongside the middle of the centerpost. verting the preceding formulae.

Inductance calculation Design Strategy Several methods are in common use for calculat- The general design procedure that will be fol- ing inductance: lowed is: I. From the circuit design, define the circuit pa- Discrete gap length, f.g : The magnetic path length of any core with a discrete gap consists of very rameters including inductance value L, full load high permeability magnetic core material (P, = 3000 - dc inductor current IFL, worst case ripple L!lpp, -100,000) in series with a small non-magnetic gap (P, max peak instantaneous short-circuit current limit = 1). In practice, the reluctance of the magnetic mate- Iscpk. absolute loss limit and max temperature rial is so small compared to the gap reluctance, that it rise. Worst case ripp/e is at max V1N for buck- can usually be neglected. The corrected gap dimen- derived, at min V1Nfor boost. Fu// /oad inductor sions alone determine the inductance: current equa/s /oad current for buck on/y. Refer to Section 2. L = .uON2 A;;{. X 10-2 Henrys (3a) 2. Select the core material. Refer to Section 2. (Sl units, dimensions in cm) 3. Determine the maximum nux density and Ag = corrected gap area (page 5-4) max. nux swing at which the core will be oper- Effective permeability, 'Ue: Whether the gap is ated (limited either by saturation or by core loss). discrete or distributed, it is a small total length of Defme a conservative saturation limit, BMAX(per- haps 3000Gauss (0.3Tesla) for power ferrite). If non-magnetic material in series with a much greater length of high permeability magnetic material. The the core is saturation limited, B MAX will be reached at Iscpk. With a discrete gapped core, the actual core can be considered equivalent to a solid gap has the most significant innuence on the B-H homogeneous core with the same overall core dimen- characteristic, linearizing it until well into satu- sions, made entirely of an imaginary material with permeability Jie , which typically ranges from 10 (for ration. Therefore, if the core is saturation limited, a large gap) to 300 (for a small gap). This concept is maximum nux swing LiEMAX will be in the same most useful for distributed gap powdered metal cores, proportion to L!lppas BMAXisto Iscpk: where the total gap cannot be physically measured. MMAX = BMAX- Mpp (4) L = .UO.UeN2AY;e X 10-2 Henrys (3b) Iscpk (SI units, dimensions in cm)

5-7 dip dB Divide the calculated L1BMAX value by two to con- E = N- = NAe- Faraday's Law vert peak-peak flux density to peak, and enter the dt dt core loss curve (Fig. 2-3) on the "flux density" axis (really peak flux density). At the ripple fre- E=L~ quency curve, find the resulting core loss. If the M core loss is much less than 100 mw/cm3, this con- Combining the above (L in I!H, Ae in em): firms that the core is probably saturation limited, N = LM MAX x 10-2 and the calculated L1BMAX value is probably valid. (5) But if the indicated core loss is much greater than MMAXAe loo mw/cm3, the core will probably be loss lim- N must then be rounded to an integer value. If N ited. L1BMAX must then be reduced to achieve an is rounded down to a smaller integer, the core acceptable core loss (Step 5). If the core is loss may saturate, or, if the core is loss limited, core limited, peak flux density at IsGpkwill then be less loss will be greater than planned. However, than BMAX. winding loss will be reduced. If N is rounded up The approach taken above equates flux density to a larger integer value, core loss will be re- values with currents, based on the presumption of duced, but winding loss increased. When N is a a linear core characteristic. This presumption is small number of turns, there is a very large in- quite valid for a gapped ferrite or laminated metal crease in winding loss when rounding up vs. core. On the other hand, powdered metal cores rounding down. It may be advantageous to round are quite non-linear over a substantial portion of down to the smaller integral N value if the re- their range. But in high frequency switching duced winding loss outweighs the increased core applications, these cores will usu- loss. ally be limited by core loss to well below satura- When an inductor has multiple windings, the tion flux density, where is much better. lowest winding with the fewest turns usu- Nevertheless, the determination of core loss and ally dominates the rounding decision. De- max. permissible flux swing are best accom- optimization caused by rounding sometimes plished by methods defined by the manufacturers forces the need for a larger core. It may be desir- of these cores. (Also, be aware that the perme- able to alter the turns ratio, or use a smaller in- ability quoted by the manufacturer may not apply ductance value (resulting in greater ripple cur- at the conditions of the application. ) rent) to avoid increased losses or the need for a larger core. 4. Tentatively select the core shape and size. Equation 5 can be applied to any winding pro- Inexperienced designers should use the area vided N, L, and L1l are all referred to that wind- product formulae (Eq. 2a, 2b), or manufacturer's guidance. Record the important core dimensions. ing. After the integer value of N has been established, 5. Determine loss limit. First, define the thermal recalculate LIB, inverting Eq.5, and determine the resistance from the data sheet or calculate RT ac- resulting core loss. cording to page 4-2. Divide the temperature rise limit by the thermal resistance to calculate the 7. Calculate the gap length required to achieve the temperature rise loss limit. Compare the tem- required inductance, using the N value estab- perature rise loss limit with the absolute loss lished in Step 6 (inverting the inductance for- limit, and use whichever is smaller. mula: Eq. 3a, 3b). If the core is limited by loss rather than by satu- For a discrete gap, the effective magnetic path ration, initially apportion half of the loss limit to length is the gap, £g. The center-pole area, Ae the core and half to the windings. Then apply the must be corrected for the fringing field (Eq. la or core loss limit to the core loss curves to fmd the I b) to obtain the effective gap area Ag. A L1BMAX value that will produce that core loss. f 9 = .uON2 --.!. X 104 (6a) 6. Calculate the number of turns, N, that will pro- L vide the desired inductance value when operated at the max flux density swing as defined in Step 3 or Step 5.

5-8 transfonnerturns ratio): VINRange: 13.33- 25.33V Output1: 5 V FullLoad Current, IFL: 50 A CircuitTopology: Forward Converter SwitchingFreq, Is: 200kHz MaxDuty Cycle: .405 (at MinVIN) MinDuty Cycle: .213 (at MaxVIN) MaxRipple Current, L1lpp: 50A x 20%= 10A Maxpeak Current, IsCpk: 65 A Inductance,L: 2.2~ H MaxLoss (absolute): 2.5 W MaxoC Rise: 40°C CoolingMethod: Natural Convection

2: Select the core material, using guidance from the L£ .u = e -4 manufacturer's data sheet. e (7)

.u N2 A 0 e Core Material: Ferrite, MagneticsType p (L in ~H, dimensions in cm) 3. Determine max. flux density and max. flux swing 8. Calculate the conductor size and winding re- at which the core will be operated. A saturation- sistance. (Refer to the following cookbook sec- limited BMAXofO.3T (3000 Gauss) will be used. If tions for details. ) the core is saturation limited, it will be at the Winding resistance is calculated using the con- BMAXlimit when the peak current is at the short- ductor cross-section area and length. circuit limit. The max. peak-peak flux density Resistivity of copper: swing corresponding to the max. current ripple Pcu = 1.724[1+.0042(T-20)]X10-6 Q-cm will then be:

M 10 Pcu = 2.30x 10-6 Q-cm at lOO°C ABMAX = BMAX ~ = 0.3- =.046 Tesla Iscpk 65 dc winding resistance: Dividing the peak-peak flux density swing by 2, Rx = Pcu l;/Ax Q (8) the peak flux swing is .023T (230 Gauss). Enter- (Dimensions in cm) ing the core loss curve for type P material (page 2-5) at 230 Gauss, and at the 200kHz ripple fre- 9. Calculate winding loss, total loss, and tem- quency, the core loss is approximately 4mw/cmJ. perature rise. If loss or temperature rise is too This is so much less than the loo mw/cmJ rule of high or too low, iterate to a larger or smaller core thumb that core loss will be almost negligible, size. and core operation will be saturation limited at The cookbook design examples which follow will Iscpk.The maximum flux density swing, ABMAX,is more fully illustrate this design process. therefore .046T as previously calculated.

Cookbook Example: 4. Tentatively select core shape and size, using guidance from the manufacturer's data sheet or Buck Output Filter Inductor using the area product formula given previously. In Section 4, a forward converter transformer was Core type, Family: E-E core -ETD Series designed for 5V, 50A output. This filter inductor will be designed as the output filter for this same Using the saturation limited Area Product for- power supply. mula, with BMAX= 0.3T and Kl =.03, an Area Product of 0.74 cm4 is required. An ETD34 core I. Define the power supply parameters pertaining to size will be used, with AP = 1.21 cm4 (with bob- the inductor design. (Vin for the inductor equals Vin for the transformer divided by the 7.5:1 bin).

5-9 Core Size: 34mm --ETD34 8. Calculate the conductor size, winding resistance, For the specific core selected, note: losses, and temperature rise. From Step 4, window breadth, bw = 2.10cm, and Effective core Area, Volume, Path Length, cen- height, hw = 0.60cm. The winding will consist of ter-pole diameter (cm): 5 turns (5 layers) of copper strip, 2.0cm wide, Ae: 0.97 cm2 spiral wound, with .05mm (2 mil) low voltage in- V.: 7.64 cm3 sulation between layers. £.: 7.9 cm At 50A full load current, 450 Alcm2 requires a Dcp: 1.08 cm conductor area of 0.111 cm2, Dividing this con- Window Area, Breadth, Height, Mean Length per ductor area by the 2.0cm width requires a thick- Turn (with bobbin): ness of .0555cm. Five layers, including .005cm Aw: 1.23 cm2 insulation between layers, results in a total bw: 2.10cm winding height of 0.3cm, half the available win- hw: 0.60 cm dow height. MLT: 6.10cm To reduce losses, increasing copper thickness to 5. Define RT and Loss Limit. Apportion losses to the 0.1cm results in a total winding height of .525cm, core and winding. Thermal resistance from the and a conductor area of 0.2cm2. Five turns with a data sheet is 19°C/Watt. Loss limit based on max. mean length/turn = 6.1cm results in a total wind- temperature rise: ing length of 30.5cm. Winding resistance: Plim = °Crise/RT: = 40/19 = 2.1 Watts 1 ~ 30.9 Rdc = p- = 2.3 X 10 .-=.000355.0. Compared to the absolute loss limit of 2.5W A 0.2 (Step I ), the temperature rise limit of 2.1 W ap- plies. Core loss is 4 mW/cm3 (Step 3): DC loss: 502..000355 = 0.89 Watts Pc = mW/cmJ xV.=4 x 7.64 = 30mW With reference to Section 3-4, the ac loss is cal- culated. Skin depth DpEN= .O17cm at 200kHz. Therefore, winding loss can be as much as With a conductor thickness of 0.1cm, Q = 2 Watts. However, since the core is larger than 0.1/.017 = 5.9. Entering Dowell's curves, page 3- the Area Product calculation suggests, it should 4, with Q=6 and 5 layers, RAc/RDc is approxi- be possible to reduce the winding loss. mately loo, so that RAc = .035Q. 6. Calculate the number of turns that will provide The rms value of the triangular ripple current the desired inductance value: waveform equals L1Ipp/..J12 Since max L1Ippis N= LMMAX x10-2 (5) 10A, Irms = 10/..J12= 2.9A. MMAXAe Therefore the ac loss is: (L in I!H, dimensions in cm) PLac= rR = 2.92..035 = 0.29 Watts

2.2.10 2 N = x10- = 4.93 ~ 5 Turns Total winding loss dc plus ac is: .046.0.97 Pw = 0.89 + 0.29 = 1.18 Watts

7. Calculate the gap length that will achieve the re- 9. Since the core loss is only 30mW, the total loss, quired inductance value: 1.21W, is considerably less than the 2.1W limit I-g =.UoN ~ ( 1+-L 1- ) 2 originally calculated. The windings operate with 2 A only 250 Alcm2 at full load, accounting for the XlO4 (6b) L Dcp reduced loss. This is because the ETD34 core has an Area Product 65% greater than the calculated (L in ~H, dimensions in cm) requirement. A smaller core could possibly have 2 been used. However, the ETD34 core provides 1 £g +- ) XlO4 improved power supply efficiency. 1.08

J!g = 0.192 cm

5-10 Worst case for core saturation is at maximum Coupled Filter Inductors Buck-derived converters with multiple outputs peak current, occurring at low line voltage at the peak often use a single filter inductor with multiple cou- of the rectified line voltage waveform. This is usually easy to calculate, regardless of the operating mode. pled windings, rather than individual inductors for each output. The design process for a coupled induc- Note that the simple boost topology has no inher- tor, discussed in Ref. R5 is essentially the same as ent current limiting capability, other than the series for a single-winding inductor. resistance of the line, , and bulk filter ca- The process is simplified by assuming that all pacitor. Thus, the boost inductor will saturate mo- windings are normalized and combined with the low- mentarily during start-up, while the bulk capacitor est voltage winding. Finally, divide the copper cross- charges. The resulting is basically the section area into the actual multiple windings. Each same as with a simple capacitor-input filter, and is winding will then occupy an area (AwN) proportional usually acceptable in low power applications. In high to its power output. power applications, additional means of inrush cur- rent limiting is usually provided -a or an Boost Inductor input buck current limiter. While saturation may be permissible during startup, the circuit must be de- In a simple boost application, the inductor design signed so that the inductor does not saturate during is essentially the same as for the dis- worst-case normal operating conditions. cussed previously. Calculating the losses averaged over the rectified In switching power supplies, boost topologies are line period is a difficult task. Averaged losses can be widely used in Correction applications approximated by assuming VIN is constant, equal to and in low voltage battery power sources. Otherwise, the rms value of the actual rectified line voltage the boost configuration is rarely used. waveform. In a PFC application, boost inductor design is Because input current is greatest at low input complicated by the fact that the input voltage is not voltage, low frequency winding losses are greatest at dc, but the continuously varying full-wave rectified low VIN. line voltage waveform. Thus, as VIN changes with the For discontinuous mode operation, L1lp-pis great- line voltage wavefonn, the high frequency waveforms est at low VIN. Therefore, core loss and ac winding must also change. High frequency ripple current, flux loss are worst case at low line. swing, core loss and winding loss all change radically However, when the boost topology is operated in throughout the rectified line period. the continuous mode, max L1lp-pand worst case core The situation is further complicated by the fact that in different PFC applications, the boost topology loss and ac winding loss occur when V1Nequals one- may be designed to operate in one of a wide variety half of V0. But since L1lp-pis usually very much of modes: smaller than low frequency current, core loss and ac winding loss are usually negligible in continuous .Continuous mode, fIXed frequency mode operation. .Continuous mode, variable frequency Because the boost inductor design follows the .At the mode boundary, variable frequency same pattern as the output filter inductor design pre- .Discontinuous mode, fIXed frequency viously covered, a cookbook example of boost in- ductor design is not given. It is left to the designer to .Discontinuous mode, variable frequency determine the worst-case current values governing the .Continuous mode, transitioning to discon- design. tinuous during the low current portion of the line cycle, and at light loads. Flyback Transformer Design As in the buck-derived applications, the limiting Figures 5-6 and 5-7 show the inductor current factors for the boost inductor design are (a) losses, waveforms for continuous mode and discontinuous averaged over the rectified line period, and/or (b) mode operation. All currents are normalized to their core saturation at maximum peak current. ampere-turn equivalent by multiplying primary and secondary currents by their respective winding turns. The ampere-turns driving the core are thus propor- tional to the normalized currents shown.

5-11 The design of the flyback transformer and calcu- lation of losses requires definition of duty cycle, D, from which the transformer turns ratio, n, is calcu- lated according to the relationship: For trapezoidal waveforms, Equation II can be YIN D n~ O' simplified by ignoring the slope of the waveform top. n=-- .D= (9) If Alpp = 0.5 Ipa, the error is only 1%. If Alpp = Ipa, the Yo' I-D' YIN +nYo' error is 4%. where Va' equals output voltage plus , switch and IR drops referred to the secondary. The lrms = FI:' (lla) above relationship applies to discontinuous mode op- eration, and for discontinuous mode operation only at For triangular waveforms, Eq. II becomes: the mode boundary ~ Theoretically, a transformer-coupled flyback cir- lrms=v31pk (llb) cuit can function with any turns ratio, regardless of V1Nor Va. However, it functions best, avoiding high For all waveforms: peak currents and , when n is such that D is approximately 0.5. (for discontinuous mode opera- lac = ~lrms2 -ldc2 (12) tion, at the mode boundary.) Circuit considerations and device ratings may dictate a turns ratio that re- Switching transition times are governed by sults in a duty cycle other than 0.5. The turns ratio transformer as well as by transis- determines the trade-off between primary and secon- tor and rectifier switching speeds. Thus it is impor- dary peak voltages and peak currents. For example, tant to minimize leakage inductance by using cores reducing n reduces the duty cycle, reduces peak with long, narrow windows, and by interleaving the switch voltage and peak rectifier current, but in- windings. Although switching transitions result in creases peak switch current and peak rectifier volt- switch and rectifier losses, they have little effect on age. transformer loss. Waveform Definitions: Before flyback trans- former design can be completed, the dc, ac and total Continuous Mode Operation rms current components of each waveform must be With continuous mode operation, core loss is calculated. Current values must be calculated at each usually not significant because the ac ripple compo- of the differing worst-case conditions relevant to core nent of the total inductor ampere-turns is small com- saturation, core loss and winding loss. pared with the full load dc component. But currents Dp Duty Cycle, primary (switch) waveform in the individual windings switch on and off, transfer- Ds Duty Cycle, secondary ( ) waveform ring ampere-turns back and forth from primary to Ds = (l-Dp) In the continuous mode and at the dis- secondary(s), as shown in Fig. 5-6. This results in very large ac current components in the windings that continuous mode boundary. will likely result in significant high frequency wind- Ipk Ripple current max peak Imin Ripple current ruin peak ing losses. The secondary current dc component is equal to &pp pk-pk Ripple current, Ipk -Imin output current, regardless of V1N.At low V1Nthe pri- lpa average value of trapezoidal peak: lpa = (Ipk + Imi.)/2. mary dc and peak currents and the total inductor am- pere-turns are greatest. Thus, the worst-case condi- The following equations can be used to calculate tion for core saturation and winding losses occurs at the dc, rms, and ac values of trapezoidal wavefonns low V1N. (continuous mode operation -Fig. 5-6). They also On the other hand, the ac ripple component of the apply to triangular wavefonns (discontinuous mode - total inductor ampere-turns, and thus core loss, is Fig. 5-8), by setting Iminto zero: greatest at high V1N.But since core loss is usually (I pk + Imin) negligible with continuous mode operation, this has Idc =D~ =D.lpa (10) little significance. 2

5-12

~ Cookbook Example (Continuous Mode): I. Define the power supply parameters pertain- ing to the flyback transfornler design.

VIN: 28 ;t 4 V Output 1: 5 V Full Load Current, IFL: 10 A Circuit Topology: Flyback, Continuous Mode Switching Freq, fs: 100 kHz Desired Duty Cycle: 0.5 at 28V input Max Ripple Current, L1lpp:5 A @ 32V (secondary) Peak Shorl-circuit Current: 25A (secondary) Secondary Inductance, L: 6.8 IJH (0=0.5, L1lpp= 5A) Max Loss (absolute): 2.0 W Max Temperature Rise: 40°C Cooling Method: Natural Convection

Preliminary Calculations: The turns ratio can be defined at nominal V1N = 28V and the de- sired duty cycle of 0.5:

n=~~=~ ~=5 Figure 5-6 Flyback Waveforms, Continuous Vo' I-D 5+0.6 1-0.5 Dividing the peak-peak flux density swing by 2, Before calculating winding losses, worst-case dc the peak flux swing is .03T (300 Gauss). Entering and ac current components, occurring at low V1N, the core loss curve for type P material (page 2-5) must be defined. First, the duty cycle, Dp is de- at 300 Gauss, and at 100kHz ripple frequency, filled at low V1N: the core loss is approximately 2.6 mw/cmJ. This is so much less than the 100 mw/cmJ rule of Dp24 = nVo' - 5(5+0.6) V1N +nVO' -.=0.538 thumb that core loss is negligible. Thus, BMAXis 24 + 5(5 + 0.6) saturation limited at Iscpk of 25A, and ABMAX, is DS24 = 1- Dp24 = 0.462 limited to only .06T, corresponding to L1lp-pof 5Amp. Because the duty cycle and the turns ratio could 4. Tentatively select the core shape and size, using possibly be changed to optimize the windings, current guidance from the manufacturer's data sheet or calculations are deferred until later. using the area product formula. 2: Select the core material, using guidance from the Core type, Family: E-E core -ETD Series manufacturer's data sheet. Using the saturation limited Area Product for- Core Material: Ferrite, MagneticsType p mula, with BMAX= 0.3T and Kl =.0085, an Area Product of 1.08 cm4 is required. An ETD34 core 3. Determine max. flux density and max. flux swing size will be used, with AP = 1.21 cm4 (with bob- for core operation. A saturation-limited BMAXof 0.3T (3000 Gauss) will be used. If the core is bin). saturation limited, B will reach B MAXwhen peak Core Size: 34mm 00ETD34 current reaches the short-circuit limit. Assuming For the specific core selected, note: reasonable linearity of the gapped core B-H char- acteristic, L!BMAXwith max. current ripple (at Effective core Area, Volume, Path Length, Cen- 32V) will be: ter-pole diameter (cm): As: 0.97 cm2 Mpp 5 Ve: 7.64 cm3 MMAX =BMAX-=0.3-=.06 Tesla (4) I SCpk 25 £e: 7.9 cm Dcp: 1.08 cm

5-13

~ Window Area, Breadth, Height, Mean Length per Secondary Side- VIN= 24V, Ds= 0.462 Turn (with bobbin): (Eq.10, 11a, 12) Aw: 1.23 cm2 bw:2.10cm Output dc Current, Is", : 10 A hw: 0.60 cm I 10 Avg peak Current, Ispa: J!;£ = = 21.65A MLT: 6.10cm Ds .462 5. Defme RT and Loss Limit. Apportion losses to the core and winding. Thermal resistance from the rmsCurrent, Isnns : ,[ii;:[;' = 14.7 A core data sheet is 19°C/Watt. Loss limit based on max. temperature rise: ac Current, Isac: ~ Irms2 -IdC2 = 10.77 A Plim = °Crise/RT: = 40/19 = 2.1 Watts The secondary consists of 6 turns (6 layers) of Since this exceeds the 2.0W absolute loss limit copper strip, 1.5cm wide and .015 cm thick, spi- from Step I, The 2.0W limit applies. ral wound. Conductor area is .015xl.5 = .0225 cm2. Current density is 14.7N.0225 = 650 a/cm2. Core loss is: Pc = mW/cm3 x V. = 2.6 x 7.64 = 20mW Six layers, including .005cm (2 mil) low voltage insulation between layers, results in a total Therefore, core loss is negligible. The entire 2.0 winding height of 0. 12cm. Watt loss limit can be allocated to the winding. Six turns with mean length/turn = 6.1cm results 6. Calculate the number of secondary turns that will in a total winding length of 36.6cm. Winding re- provide the desired inductance value: sistance: N= LMMAX XlO-2 (5) AB MAX Ae

(L in ~H, dimensions in cm) Calculating ac resistance: DpEN at 100kHz = .024cm. With a conductor thickness of .015cm, Q 6.8.5 -2 =.015/.024 = 0.625. Entering Dowell's curves, N.~ = xlO =5. 84 ~ 6 Turns .06.0.97 page 3-4, with Q = 0.625 and 6 layers, RAc/RDCis Np =Ns xn=6x5=30 Turns approximately 1.6. Roc = Rdc X 1.6 = .0037.0. X 1.6 =.0059.0 7. Calculate the gap length to achieve the induc. tance value with minimum N. Primary Side -V1N = 24V, Dp = 0.538 A ( £ (Eq.10, 11a, 12) £ = .u N2--.£. 1 +-L XlO4 9 O L D (6b) CP Note that the primary and secondary average peak ampere-/urns are always equal, and together (L in ~H, dimensions in cm) constitute the dc ampere-turns driving the induc- tor core. Thus the primary avg. peak current, Ippa = Isp/n. I! =41fX10-7.52~ ( 1+!J- xlO4 9 6.8 1.08 avg. peak Current, /ppa: Ispafn = 21.6% = 4.33A £g =.080 cm dc Current, /Prk:: D. I Ppa = 0.538 x 4.33 = 2.33A 8. Calculate the conductor sizes and winding resis- tances: rms Current, /Prms: ~ = 3.18A From Step 4, window breadth, bw = 2.l0cm, and ac Current, /Pac: ~ Irms2 -Idc2 = 2.16A height, hw = 0.60cm. A creepage allowance of 0.3 cm is necessary at each end of the windings. Winding width is 2.l0cm minus (2xO.3) = 1.5cm. peak SC Current, Iscpk:251n = 5 A

5-14 The primary winding consists of 30 turns of Litz The available winding height could permit a with OD of 0.127cm, in three layers, 10 thicker secondary conductor. This would reduce turns in each layer. OD enables 10 turns dc loss, but the resulting increase in ac loss be- to fit across the 1.5cm winding breadth. The cause of the larger Q value would exceed the dc height of the 3-layer primary is 3xO.127 = loss reduction. 0.381cm. Primary dc loss (Rdc= .0567.0.):

The Litz wire consists of 150 strands of #40A WG PPdc = I Pdc2Rdc = 22.0567 = 0.225 Watts wire (OD=.0081cm). From the wire tables, #40A WG wire has a resistance of .046 .Q/cm di- Primary ac loss (Rac= .090.0.): vided by 150 strands, resulting in a resistance of PPac = Ipac2 Rac = 2.162x.090 = 0.42 W .00031 .Q/cm at 100°C. The wire length equals 30 turns times ML T of Total primary winding loss --dc plus ac is: 6.1cm= 183cm. Ppw= 0.225 + 0.42 = 0.645 Watts

Primary winding resistance: Total winding loss: Rdc =.00031 Q / cm x 183cm =.0567 Q Pw = 1.05 + 0.645 = 1.695 Watts

To calculate the ac resistance, the 150 strand Since the core loss is only 20m W, the total loss, #40A WG Litz wire is approximately equivalent 1.7IW, is within than the 2.0W absolute loss to a square array 12 wide by 12 deep (square root limit. of 150 ). There are therefore a total of 36 The total winding height, including .02cm isola- layers of#40AWG wire (3layers times 12). tion: 0.12 + 0.381 + .02 = 0.521cm, within the avail- Center-to-center spacing of the #40A WG wires able winding height ofO.60cm. equals the winding width of 1.5cm divided by Mutual inductance of 6.8J.1.Hseen on the secon- 120 (10 Litz wires times 12 wide #40A WG wires dary winding translates into 170J.1.Hreferred to the within the Litz), a spacing, s, of .0125cm. primary (Lp=n2Ls). From Reference R2, pg 9, the effective layer Leakage inductance between primary and secon- thickness equals: dary calculated according to the procedure presented in Reference R3 is approximately 5J.1.H,referred to the primary side. Interwinding capacitance is ap- proximately 50pF . =.0054cm If the windings were configured as an interleaved Therefore, referring to Figure 3-5, structure (similar to Figure 4-1), leakage inductance will be more than halved, but interwinding capaci- Q =.0054cm / DpEN =.0054/.024 =.225 tance will double. The interleaved structure divides and with 36 layers, RAcIRDC= 1.6, and the winding into two sections, with only half as many layers in each section. This will reduce Ra/Rdc to Rac = Rdc X 1.6 =.0567 X 1.6 =.0900 nearly 1.0 in both primary and secondary, reducing ac losses by 0.35W, and reducing the total power loss 9. Calculate winding loss, total loss, and tempera- from 1.71W to 1.36W. The secondary copper thick- ture rise: ness could be increased, further reducing the losses. Secondary dc loss: Discontinuous Mode Operation Psdc=Idc2Rdc=102..0037=0.37 Watts Discontinuous mode waveforms are illustrated in Figure 5-7. By definition, the total ampere-tums Secondary ac loss: dwell at zero during a portion of each switching pe- PSac = ISac2 .Rac = 10. 7r..0059 = 0.68W riod. Thus, i~ the discontinuous mode there are three distinct time periods, tT, tR, and to, during each Total secondary winding loss --dc plus ac is: switching period. As the load is increased, peak cur- Psw= 0.37 + 0.68 = 1.05 Watts rents, tT, and tR increase, but to decreases. When to

5-15 becomes zero, the mode boundary is reached. Further Cookbook Example (Discontinuous Mode): increase in load results in crossing into continuous I. Define the power supply parameters pertaining to mode operation. This is undesirable because the con- the flyback transformer design. trolloop characteristic suddenly changes, causing the control loop to become unstable. VIN: 28 j; 4 V In the discontinuous mode, all of the energy Output 1: 5 V stored in the inductor (Y2Llpk2)is delivered to the out- Full Load Current, IFL: 10 A put during each cycle. This energy times switching Current: 12 A frequency equals output power. Therefore, if fre- Circuit Topology: Flyback, Discontinuous quency, L, and Vo are held constant, Y2Llpk2does not Switching Freq, Is: 100 kHz vary with VIN,but is proportional to load current only, Desired Duty Cycle: 0.5 at 24V, mode boundary and Ipk is proportional to the square root of load cur- Estimated ISCpk:45 A (secondary) rent. However, VIN,n, D and L collectively do deter- Est. Sec. Inductance: 0.6211H (D=0.5, ~1p-p=45A) mine the maximum stored energy and therefore the Max Loss (absolute): 2.0 W maximum power output at the mode boundary. Max Temperature Rise: 40°C The circuit should be designed so that the peak Cooling Method: Natural Convection short-circuit current limit is reached just before the Preliminary Calculations: The turns ratio is mode boundary is reached, with turns ratio, duty cy- defined based on min V1N (24V) and Vo ' (5.6V) cle and inductance value designed to provide the nec- and the desired duty cycle of 0.5 at the mode essary full load power output at a peak current less than the current limit. boundary: The circuit design can never be completely sepa- n = ~ ~ = ~ ~ = 4.28 ~ 4 rated from the design of the magnetic components. Vo' 1-D 5+0.6 1-0.5 This is especially true at high frequencies where the small number of secondary turns can require difficult Turns ratio, n, is rounded down to 4: 1 rather than up choices. The ideal design for a discontinuous mode to 5:1 because: (a) 4:1 is closer, (b) peak output cur- flyback transformer might call for a secondary with rent is less, reducing the burden on the output filter 1YO turns. The choice of a 1 turn or 2 turn secondary capacitor, and ( c) primary switch peak voltage is less. may result in a size and cost increase, unless the The duty cycle at the mode boundary is no longer 0.5, turns ratio and duty cycle are changed, for example. and must be recalculated:

v '.n o 5.6x4 Dp24 = = 0.483

V1N +VO'.n 24+5.6x4

A-t DS24 = 1- Dp24 = 0.517

The peak secondary current at the mode boundary is:

The peak short-circuit current limit on the pri- A-t mary side should therefore be set slightly below 11.6A (=46.4/n). The inductance value required for the secondary current to ramp from 46.4A to zero at the mode 0, IT tR 10 boundary is:

FiRUre 5-7 Discontinuous Waveforms

5-16

~ As: 0.56 cm2 Ve: 3.48 cm3 fe: 6.19 cm L = 5.6~ = 0.624 .uH Dcp: 0..85 cm 46.4 Window Area, Breadth, Height, Mean Length per Before calculating winding losses, it is necessary Turn ( , indicates reduced dimensions with bob- to define worst-case dc and ac current compo- bin): nents, occurring at low V1N.Because the turns ra- Aw I Aw': 1.02/0.45 cm2 tio and the duty cycle could possibly be changed bwlbw': 2.07 11.72cm to optimize the windings, current calculations hw I hw': 0.50 10.38 cm will be deferred until later . MLT: 4.63 cm 2: Select the core material, using guidance from the 5. Define RT and Loss Limit, and apportion losses to manufacturer's data sheet. the core and winding. Thermal resistance from Core Material: Ferrite, MagneticsType p the core data sheet is 28°C/Watt. Loss limit based on max. temperature rise: 3. Determine max. flux density and max. flux swing at which the core will be operated. A saturation- Plim = °Crise/RT: = 40/28 = 1.42 Watts limited BMAXofO.3T (3000 Gauss) will be used. Since this is less than the 2.0W absolute loss In the discontinuous mode, the current is zero limit from Step 1, The 1.42W limit applies. during a portion of each switching period, by definition. Therefore & always equals Ipeakand Preliminary core loss calculation: since they are proportional, LiB must equal Bpeak. Pc = mWlcmJ x Ve = 100 X 3.48 = 350 mW LiBMAXandBMAXOCCur at low V1Nwhen the current 6. Calculate the number of secondary turns that will is peak short-circuit limited. To determine if provide the desired inductance value: LiBMAX is core loss limited, enter the core loss curve for type P material at the nominal N = LM MAX X 10-2 (5) 100mw/cm3loss limit, and at 100kHz ripple fre- ABMAXAe quency, the corresponding max. peak flux density is 1100 Gauss. Multiply by 2 to obtain peak-peak (L in ~H, dimensions in cm) flux density swing LiBMAXof2200 Gauss, or 0.22 0.63.46 -2 N s = X 10 = 2.35 ~ 2 Turns Tesla. Since in the discontinuous mode, BMAX .22.0.56 equals LiBMAX , then B MAXis also limited to 0.22T , Np =Ns xn=4x2=8 Turns well short of saturation. Thus, in this application, Because Ns was rounded down from 2.35 to 2 the core is loss-limited at LiBMAX=0.22T, corre- sponding to &p-p = Iscpk= 46A. turns, flux density swing is proportionately greater than originally assumed: 4. Tentatively select the core shape and size, using 2.35 MMAX = 0.22- = 0.258 guidance from the manufacturer's data sheet or Tesla 2 using the area product formula. Divide by 2 to obtain peak flux density swing and Core type, Family: E-E core -ETD Series enter the core loss curve at 0.13T (1300 Gauss) to obtain a corrected core loss of 160mW/cmJ Mul- Using the loss limited Area Product formula, with tiply by Ve = 3.48 cm for a corrected core loss of LiBMAX=0.22T and K2 =.006, an Area Product of 560 mW. 0.31 cm4 is required. An ETD24 core size will be If Ns had been rounded up to 3 turns, instead of used, with AP = 0.37 cm4 (with bobbin). down to 2 turns, core loss would be considerably Core Size: 24mm -ETD24 less, but winding loss would increase by an even greater amount, and the windings might not fit For the specific core selected, note: into the available window area.. Effective core Area, Volume, Path Length, Cen- 7. Calculate the gap length to achieve the induc- ter-pole diameter (cm): tance value:

5-17 Two turns --two layers, including .005cm (2 (6b) mil) low yoltage insulation between layers, re- sults in a total winding height of .081cm. (L in ~H, dimensions in cm) Two turns with mean length/turn = 4.6 cm results in a total winding length of 9.2 cm.

~ xlO4 Secondary dc resistance: 0.95 )2 1 -6 9.2 f!g =.050 cm Rdc = p- = 2.3 X 10 .-=.00049.0. A .043 8. Calculate the conductor sizes and winding resis- Calculating ac resistance: DpEN at 100kHz = tances: .024cm. With a conductor thickness of .033cm, Q From Step 4, window breadth, bw = 1.72 cm, and =.038/.024 = 1.6. Entering Dowell's curves, page height, hw = 0.38cm. A creepage allowance of 0.3 3-4, with Q = 1.6 and 2 layers, RAcIRDC is ap- cm is necessary at each end of the windings. proximately 2.5. Winding width is 1.72 cm minus (2xO.3) = Secondary ac resistance (non-interleaved): 1.12cm. Roc = Rdc X 2.0 =.00049.0. X 2.5 =.00122.0. Secondary side: V1N= 24V, Ds = 0.517 (Eq.11b,12) If the windings are interleaved, forming two winding sections as shown in Fig. 5-8, there is Output dc Current, 'Sdc: 12 A (Short Circuit) one secondary turn in each section. Entering Peak S.C. Current, Ispk: 46.4 A Dowell's curves with Q = 1.6 and one layer, RAc/RDcis 1.5. rms Current, Isrms: ~3.~ I pk Secondary ac resistance (interleaved): Roc = Rdc X 1.5 = .00049.0. X 1.5 =.0007.0.

Primary side: V1N= 24V, Dp = 0.483 I 2 2 ac Current, ISac: '.JIrms -Idc = 15 A (Eq.11b,12) Note that primary and secondary peak ampere- Secondary conductor area for 450 A/cm2 requires turns are always equal. Thus the primary peak cur- 19.2A/450 = .043 cm2 (AWG 11). This is imple- rent, Ippk= Isp';n. mented with copper strip 1.12cm wide and .038 cm thickness, 2 turns, spiral wound. peak S.C.Current,/Ppk I BOBBIN

3 LAYERS

Figure 5-8- Interleaved Flyback Windings

5-18 Total loss, including core loss of 0.56 W: Eight layers, including .005cm (2 mil) low volt- age insulation between layers results in a total PT =Pw +Pc =0.42+0.56=0.98 W winding height of8x.014 = 0.112cm. The total loss is well within the 1.42 Watt limit. Eight turns with mean length/turn = 4.6 cm re- The temperature rise will be: sults in a total winding length of36.8 cm. °Crise = RT XPT = 28 x 0.98 = 27 °C Primary dc resistance: Total winding height, including two layers of 1 -6 36.8 Rdc = p- = 2.3 X 10 .-=.0085.0. .02cm isolation: 0.04 + 0.081 + .112 = 0.233 cm, well A .01 within the 0.38 cm available. Calculating ac resistance: DpEN at 100kHz = Leakage inductance between primary and secon- .024cm. With a conductor thickness of .009cm, Q dary calculated according to the procedure presented =.009/.024 = .375. Entering Dowell's curves, in Reference R3 is approximately .08JlH, referred to page 3-4, with Q = .375 and 8 layers, RAc/RDc is the primary. Interwinding capacitance is approxi- approximately 1.2. mately 50pF. Primary ac resistance (non-interleaved): If the windings were not interleaved (as shown in Figure 5-8), leakage inductance would be more than Rac = Rdc X 1.2 = .0085.0. X 1.2 =.Oill doubled, but interwinding capacitance would be With the interleaved structure, there are only 4 halved. Interleaving also helps to reduce ac winding layers in each winding section. Entering Dowell's losses. This becomes much more significant at higher curves, page 3-4, with Q = .375 and 4 layers, power levels with larger conductor sizes. RAc/Roc is 1.0. It is very important to minimize leakage induc- tance in ftyback circuits. Leakage inductance not only Primary ac resistance (interleaved): slows down switching transitions and dumps its en- Rac = Rdc = .0085.0. ergy into a clamp, but it can steal a very significant amount of energy from the mutual inductance and 10. Calculate winding loss, total loss, and tempera- prevent it from being delivered to the output. ture rise, using the interleaved structure: Secondary dc loss (Rdc= .00049.0.): References PSdc= 122..00049 = 0.07 Watts "R-numbered" references are reprinted in the Reference Section at the back of this Manual. Secondary ac loss (Roc= .0007.0.): p Sac= I Sac2 .Rac = 152..0007 = 0.16 W (R3) "Deriving the Equivalent Electrical Circuit from the Magnetic Device Physical Properties," Uni. trade Seminar Manual SEMIOOO, 1995 and Total secondary winding loss --dc plus ac is: SEMI I 00, 1996 Psw= 0.07 + 0.16 = 0.23 Watts

Primary dc loss (Rdc= .0085.0.): (R5) "Coupled Filter Inductors in Multiple Output Buck Regulators Provide Dramatic Performance lm- Ppdc= IpdC2Rdc = 2.82x.0085 = .067 Watts provement," Unitrode Seminar Manual SEMI 100, 1996 Primary ac loss (Roc= .OO85.Q): (1) MIT Staff, "Magnetic Circuits and Transfonn- Ppoc =Ipac2Roc =3.712x.OO85=0.12 Watts ers," M/T Press, 1943 Total primary winding loss --dc plus ac is: Ppw= .07 + 0.12 = 0.19 Watts

Total winding loss: Pw = 0.23 + 0.19 = 0.42 Watts

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