Chapter 13: Filter inductor design


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1 Chapter 13. Filter Inductor Design Several types of magnetic devices, their BH loops, and core vs. copper loss Filter inductor design constraints The core geometrical constant K g A stepbystep design procedure Summary of key points 1
2 13.1. Several types of magnetic devices, their BH loops, and core vs. copper loss A key design decision: the choice of maximum operating flux density B max Choose B max to avoid saturation of core, or Further reduce B max, to reduce core losses Different design procedures are employed in the two cases. Types of magnetic devices: Filter inductor Conventional transformer Flyback transformer Magnetic amplifier Ac inductor Coupled inductor SEPIC transformer Saturable reactor 2
3 Filter inductor CCM buck example L B + i(t) minor BH loop, filter inductor B sat H c i(t) H c0 H c I i L BH loop, large excitation 0 DT s T s t 3
4 Filter inductor, cont. Negligible core loss, negligible proximity loss Loss dominated by dc copper loss Flux density chosen simply to avoid saturation + v(t) i(t) n turns core reluctance R c Φ air gap reluctance R g Air gap is employed Could use core materials having high saturation flux density (and relatively high core loss), even though converter switching frequency is high n i(t) + R c Φ(t) R g 4
5 Ac inductor L B i(t) i(t) BH loop, for operation as ac inductor B sat i H c i t H c H c core BH loop 5
6 Ac inductor, cont. Core loss, copper loss, proximity loss are all significant An air gap is employed Flux density is chosen to reduce core loss A highfrequency material (ferrite) must be employed 6
7 Conventional transformer + i 1 (t) i mp (t) n 1 : n 2 i 2 (t) + B v 1 (t) L mp v 2 (t) v 1 (t) area λ 1 conventional transformer BH loop, for operation as λ 1 2n 1 A c n 1 i mp l m H c i mp (t) core BH loop i mp t H(t)= ni mp(t) l m 7
8 Conventional transformer, cont. Core loss, copper loss, and proximity loss are usually significant No air gap is employed Flux density is chosen to reduce core loss A high frequency material (ferrite) must be employed 8
9 Coupled inductor Twooutput forward converter example i 1 (t) I 1 i 1 n i v 1 i 2 (t) I 2 i 2 v g + B t n 2 turns i 2 + minor BH loop, coupled inductor v 2 H c H c0 H c BH loop, large excitation 9
10 Coupled inductor, cont. A filter inductor having multiple windings Air gap is employed Core loss and proximity loss usually not significant Flux density chosen to avoid saturation Lowfrequency core material can be employed 10
11 DCM Flyback transformer i 1 (t) n 1 : n 2 i 1 i mp i 2 + i 1,pk v g + L mp v i 2 (t) B i mp (t) t i 1,pk BH loop, for operation in DCM flyback converter n 1 i 1,pk R c l m R c +R g H c t core BH loop 11
12 DCM flyback transformer, cont. Core loss, copper loss, proximity loss are significant Flux density is chosen to reduce core loss Air gap is employed A highfrequency core material (ferrite) must be used 12
13 13.2. Filter inductor design constraints Objective: i(t) L R Design inductor having a given inductance L, which carries worstcase current I max without saturating, and which has a given winding resistance R, or, equivalently, exhibits a worstcase copper loss of P cu = I rms 2 R 13
14 Assumed filter inductor geometry + v(t) i(t) n turns Φ crosssectional area A c air gap l g R c = l c µ c A c R g = l g µ 0 A c R c Solve magnetic circuit: ni = Φ R c + R g n i(t) + Φ(t) R g For R c >> R g : ni ΦR g 14
15 Constraint: maximum flux density Given a peak winding current I max, it is desired to operate the core flux density at a peak value B max. The value of B max is chosen to be less than the worstcase saturation flux density of the core material. From solution of magnetic circuit: ni = BA c R g Let I = I max and B = B max : ni max = B max A c R g = B max l g µ 0 This is constraint #1. The turns ratio n and air gap length l g are unknown. 15
16 Constraint: Inductance Must obtain specified inductance L. We know that the inductance is L = n2 R g = µ 0 A c n 2 l g This is constraint #2. The turns ratio n, core area A c, and air gap length l g are unknown. 16
17 Constraint: Winding area Wire must fit through core window (i.e., hole in center of core) core Total area of copper in window: na W Area available for winding conductors: wire bare area A W core window area W A K u W A Third design constraint: K u W A na W 17
18 The window utilization factor K u also called the fill factor K u is the fraction of the core window area that is filled by copper Mechanisms that cause K u to be less than 1: Round wire does not pack perfectly, which reduces K u by a factor of 0.7 to 0.55 depending on winding technique Insulation reduces K u by a factor of 0.95 to 0.65, depending on wire size and type of insulation Bobbin uses some window area Additional insulation may be required between windings Typical values of K u : 0.5 for simple lowvoltage inductor 0.25 to 0.3 for offline transformer 0.05 to 0.2 for highvoltage transformer (multiple kv) 0.65 for lowvoltage foilwinding inductor 18
19 Winding resistance The resistance of the winding is R = ρ l b A W where ρ is the resistivity of the conductor material, l b is the length of the wire, and A W is the wire bare area. The resistivity of copper at room temperature is Ωcm. The length of the wire comprising an nturn winding can be expressed as l b = n (MLT) where (MLT) is the meanlengthperturn of the winding. The meanlengthperturn is a function of the core geometry. The above equations can be combined to obtain the fourth constraint: R = ρ n (MLT) A W 19
20 13.3 The core geometrical constant K g The four constraints: ni max = B max K u W A na W l g L = µ 0A c n 2 µ 0 l g These equations involve the quantities R = ρ n (MLT) A W A c, W A, and MLT, which are functions of the core geometry, I max, B max, µ 0, L, K u, R, and ρ, which are given specifications or other known quantities, and n, l g, and A W, which are unknowns. Eliminate the three unknowns, leading to a single equation involving the remaining quantities. 20
21 Core geometrical constant K g Elimination of n, l g, and A W leads to A 2 c W A (MLT) ρl2 2 I max 2 B max RK u Righthand side: specifications or other known quantities Lefthand side: function of only core geometry So we must choose a core whose geometry satisfies the above equation. The core geometrical constant K g is defined as K g = A c 2 W A (MLT) 21
22 Discussion K g = A 2 cw A (MLT) ρl2 2 I max 2 22 B max RK u K g is a figureofmerit that describes the effective electrical size of magnetic cores, in applications where the following quantities are specified: Copper loss Maximum flux density How specifications affect the core size: A smaller core can be used by increasing B max use core material having higher B sat R allow more copper loss How the core geometry affects electrical capabilities: A larger K g can be obtained by increase of A c more iron core material, or W A larger window and more copper
23 13.4 A stepbystep procedure The following quantities are specified, using the units noted: Wire resistivity ρ (Ωcm) Peak winding current I max (A) Inductance L (H) Winding resistance R (Ω) Winding fill factor K u Core maximum flux density B max (T) The core dimensions are expressed in cm: Core crosssectional area A c (cm 2 ) Core window area W A (cm 2 ) Mean length per turn MLT (cm) The use of centimeters rather than meters requires that appropriate factors be added to the design equations. 23
24 Determine core size K g ρl2 2 I max (cm 5 ) RK u B max Choose a core which is large enough to satisfy this inequality (see Appendix 2 for magnetics design tables). Note the values of A c, W A, and MLT for this core. 24
25 Determine air gap length l g = µ 2 0LI max (m) A c B max with A c expressed in cm 2. µ 0 = 4π107 H/m. The air gap length is given in meters. The value expressed above is approximate, and neglects fringing flux and other nonidealities. 25
26 A L Core manufacturers sell gapped cores. Rather than specifying the air gap length, the equivalent quantity A L is used. A L is equal to the inductance, in mh, obtained with a winding of 1000 turns. When A L is specified, it is the core manufacturer s responsibility to obtain the correct gap length. The required A L is given by: A L = 10B 2 max 2 (mh/1000 turns) LI max L = A L n A c 2 (Henries) Units: A c cm 2, L Henries, B max Tesla. 26
27 Determine number of turns n n = LI max B max A c
28 Evaluate wire size A W K uw A n (cm 2 ) Select wire with bare copper area A W less than or equal to this value. An American Wire Gauge table is included in Appendix 2. As a check, the winding resistance can be computed: R = ρn (MLT) A w (Ω) 28
29 13.5 Summary of key points 1. A variety of magnetic devices are commonly used in switching converters. These devices differ in their core flux density variations, as well as in the magnitudes of the ac winding currents. When the flux density variations are small, core loss can be neglected. Alternatively, a lowfrequency material can be used, having higher saturation flux density. 2. The core geometrical constant K g is a measure of the magnetic size of a core, for applications in which copper loss is dominant. In the K g design method, flux density and total copper loss are specified. 29
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