Chapter 14: Inductor design


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1 Chapter 14 Inductor Design 14.1 Filter inductor design constraints 14.2 A stepbystep design procedure 14.3 Multiplewinding magnetics design using the K g method 14.4 Examples 14.5 Summary of key points 1
2 14.1 Filter inductor design constraints Objective: i 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 Example: filter inductor in CCM buck converter L i i I i L + 0 DT s T s t 2
3 Assumed filter inductor geometry Core reluctance R c Φ i + n v turns Air gap reluctance ni + R g F c + R c Φ R g R c = R g = l c µ c A c l g µ 0 A c Solve magnetic circuit: ni = Φ R c + R g Usually R c < R g and hence ni ΦR g 3
4 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 B sat 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. 4
5 Constraint: inductance Must obtain specified inductance L. We know that the inductance is L = n2 R g = µ 0A 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. 5
6 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 6
7 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 tha: 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 7
8 Winding resistance The resistance of the winding is R = ρ l b A W where 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 cm. 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 8
9 The core geometrical constant K g The four constraints: ni max = B max A c R g = B max l g µ 0 L = n2 R g = µ 0A c n 2 l g K u W A na W R = ρ n (MLT) A W These equations involve the quantities A c, W A, and MLT, which are functions of the core geometry, I max, B max, µ 0, L, K u, R, and, 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. 9
10 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) 10
11 Discussion K g = A 2 cw A (MLT) ρl2 2 I max 2 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 11
12 14.2 A stepbystep procedure The following quantities are specified, using the units noted: Wire resistivity ( cm) Peak winding current I max (A) Inductance L (H) Winding resistance R ( ) 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. 12
13 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 D for magnetics design tables). Note the values of A c, W A, and MLT for this core. 13
14 Determine air gap length l g = µ 2 0LI max A c B max (m) with A c expressed in cm 2. µ 0 = H/m. The air gap length is given in meters. The value expressed above is approximate, and neglects fringing flux and other nonidealities. 14
15 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. 15
16 Determine number of turns n n = LI max B max A c
17 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 D. As a check, the winding resistance can be computed: R = ρn (MLT) A w (Ω) 17
18 14.3 Multiplewinding magnetics design using the K g method The K g design method can be extended to multiplewinding magnetic elements such as transformers and coupled inductors. This method is applicable when Copper loss dominates the total loss (i.e. core loss is ignored), or The maximum flux density B max is a specification rather than a quantity to be optimized To do this, we must Find how to allocate the window area between the windings Generalize the stepbystep design procedure 18
19 Window area allocation Given: application with k windings having known rms currents and desired turns ratios v 1 = v 2 n 2 = = v k n k rms current I 1 : n 2 rms current I 2 Core Window area W A Core mean length per turn (MLT) Wire resistivity ρ Fill factor K u : n k rms current I k Q: how should the window area W A be allocated among the windings? 19
20 Allocation of winding area Winding 1 allocation α 1 W A Winding 2 allocation α 2 W A etc. { { Total window area W A 0<α j <1 α 1 + α α k =1 20
21 Copper loss in winding j Copper loss (not accounting for proximity loss) is P cu, j = I j 2 R j Resistance of winding j is R j = ρ l j A W,j with l j = n j (MLT) A W, j = W AK u α j n j length of wire, winding j wire area, winding j Hence R j = ρ n2 j (MLT) W A K u α j P cu,j = n j 2 i j 2 ρ(mlt) W A K u α j 21
22 Total copper loss of transformer Sum previous expression over all windings: P cu,tot = P cu,1 + P cu,2 + + P cu,k = ρ (MLT) W A K u k Σj =1 n 2 2 j I j α j Need to select values for 1, 2,, k such that the total copper loss is minimized 22
23 Variation of copper losses with 1 Copper loss Pcu,1 P cu,tot P cu,2 + P cu, P cu,k For 1 = 0: wire of winding 1 has zero area. P cu,1 tends to infinity For 1 = 1: wires of remaining windings have zero area. Their copper losses tend to infinity 0 1 α 1 There is a choice of 1 that minimizes the total copper loss 23
24 Method of Lagrange multipliers to minimize total copper loss Minimize the function P cu,tot = P cu,1 + P cu,2 + + P cu,k = ρ (MLT) W A K u k Σj =1 n 2 2 j I j α j subject to the constraint α 1 + α α k =1 Define the function where f (α 1, α 2,, α k, ξ)=p cu,tot (α 1, α 2,, α k )+ξ g(α 1, α 2,, α k ) g(α 1, α 2,, α k )=1 α j Σj =1 is the constraint that must equal zero and is the Lagrange multiplier k 24
25 Lagrange multipliers continued Optimum point is solution of the system of equations f (α 1, α 2, α 1, α k,ξ) =0 f (α 1, α 2, α 2, α k,ξ) =0 Result: ξ = ρ (MLT) W A K u α m = n mi m Σ n j I j n =1 k Σj =1 n j I j 2 = P cu,tot f (α 1, α 2, α k, α k,ξ) =0 f (α 1, α 2, ξ, α k,ξ) =0 An alternate form: α m = V mi m Σ V j I j n =1 25
26 Interpretation of result α m = V mi m Σ V j I j n =1 Apparent power in winding j is where V j I j V j is the rms or peak applied voltage I j is the rms current Window area should be allocated according to the apparent powers of the windings 26
27 Example PWM fullbridge transformer i 1 i 2 I turns { } n 2 turns } n 2 turns i 1 n 2 I i Note that waveshapes (and hence rms values) of the primary and secondary currents are different Treat as a threewinding transformer i 2 i 3 I 0 n 2 I 0.5I 0.5I 0 I 0.5I 0.5I 0 DT s T s T s +DT s 2T s t 27
28 Expressions for RMS winding currents i 1 n 2 I I 1 = 1 2Ts I 2 = I 3 = 1 2Ts see Appendix A 0 2T s i 2 1 dt = n 2 I 0 D 2T s i 2 2 dt = 1 2 I 1+D i 2 i 3 I n 2 I 0.5I 0.5I 0 I 0.5I 0.5I 0 DT s T s T s +DT s 2T s t 28
29 Allocation of window area: α m = V mi m Σ V j I j n =1 Plug in rms current expressions. Result: α 1 = D D Fraction of window area allocated to primary winding α 2 = α 3 = D 1+D Fraction of window area allocated to each secondary winding 29
30 Numerical example Suppose that we decide to optimize the transformer design at the worstcase operating point D = Then we obtain α 1 = α 2 = α 3 = The total copper loss is then given by P cu,tot = ρ(mlt) W A K u 3 Σj =1 n j I j 2 = ρ(mlt)n 2 2 I 2 W A K u 1+2D +2 D(1 + D) 30
31 Coupled inductor design constraints Consider now the design of a coupled inductor having k windings. We want to obtain a specified value of magnetizing inductance, with specified turns ratios and total copper loss. i 1 + v 1 i M L M : n 2 + v 2 i 2 Magnetic circuit model: R c R 1 R 2 i M + Φ R g + v k i k : n k R k 31
32 Relationship between magnetizing current and winding currents : n 2 i 1 + i M + i 2 v 1 L M v 2 R 1 R 2 Solution of circuit model, or by use of Ampere s Law: i M =i 1 + n 2 i n k i k + v k i k : n k R k 32
33 Solution of magnetic circuit model: Obtain desired maximum flux density R c Assume that gap reluctance is much larger than core reluctance: i M + Φ R g i M =BA c R g Design so that the maximum flux density B max is equal to a specified value (that is less than the saturation flux density B sat ). B max is related to the maximum magnetizing current according to I M,max = B max A c R g = B max l g µ 0 33
34 Obtain specified magnetizing inductance By the usual methods, we can solve for the value of the magnetizing inductance L M (referred to the primary winding): L M = n = n µ 0 A c R 1 g l g 34
35 Copper loss Allocate window area as described in Sectio As shown in that section, the total copper loss is then given by P cu = ρ(mlt)n2 2 1I tot W A K u with I tot = k Σ j =1 n j I j 35
36 Eliminate unknowns and solve for K g Eliminate the unknowns l g and : P cu = ρ(mlt)l 2 MI 2 2 tot I M,max 2 B max A 2 c W A K u Rearrange equation so that terms that involve core geometry are on RHS while specifications are on LHS: A 2 c W A (MLT) = ρl 2 MI 2 2 tot I M,max 2 B max K u P cu The lefthand side is the same K g as in singlewinding inductor design. Must select a core that satisfies K g ρl 2 MI 2 2 tot I M,max 2 B max K u P cu 36
37 Stepbystep design procedure: Coupled inductor The following quantities are specified, using the units noted: Wire resistivity ( cm) k n Total rms winding currents j (A) (referred to winding 1) I tot = Peak magnetizing current I M, max (A) (referred to winding 1) Desired turns ratios n 2 /. n 3 /n 2. etc. Magnetizing inductance L M (H) (referred to winding 1) Allowed copper loss P cu (W) 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) Σ j =1 I j The use of centimeters rather than meters requires that appropriate factors be added to the design equations. 37
38 1. Determine core size K g ρl 2 MI 2 2 tot I M,max (cm 5 ) P cu K u B max Choose a core that satisfies this inequality. Note the values of A c, W A, and MLT for this core. The resistivity of copper wire is cm at room temperature, and cm at 100 C. 38
39 2. Determine air gap length l g = µ 2 0L M I M,max (m) A c B max (value neglects fringing flux, and a longer gap may be required) The permeability of free space is µ 0 = H/m 39
40 3. Determine number of turns For winding 1: = L MI M,max B max A c 10 4 For other windings, use the desired turns ratios: n 2 = n 2 n 3 = n 3 40
41 4. Evaluate fraction of window area allocated to each winding α 1 = I 1 I tot α 2 = n 2I 2 I tot α k = n ki k I tot Winding 1 allocation α 1 W A Winding 2 allocation α 2 W A etc. { { 0<α j <1 α 1 + α α k =1 Total window area W A 41
42 5. Evaluate wire sizes A w1 α 1K u W A A w2 α 2K u W A n 2 See American Wire Gauge (AWG) table at end of Appendix D. 42
43 14.4 Examples Coupled Inductor for a TwoOutput Forward Converter CCM Flyback Transformer 43
44 Coupled Inductor for a TwoOutput Forward Converter v g + + i 1 v 1 Output 1 28 V 4 A f s = 200 khz n 2 turns i 2 + v 2 Output 2 12 V 2 A The two filter inductors can share the same core because their applied voltage waveforms are proportional. Select turns ratio n 2 / approximately equal to v 2 /v 1 = 12/28. 44
45 Coupled inductor model and waveforms v M + i 1 + i M i M L M v 1 I M i M Coupled inductor model : n 2 i 2 + v 2 0 v M 0 D T s V 1 Secondaryside circuit, with coupled inductor model Magnetizing current and voltage waveforms. i M is the sum of the winding currents i 1 + i 2. 45
46 Nominal fullload operating point Design for CCM operation with D = 0.35 v g + + i 1 v 1 Output 1 28 V 4 A i M = 20% of I M f s = 200 khz f s = 200 khz n 2 turns i 2 + v 2 Output 2 12 V 2 A DC component of magnetizing current is I M = I 1 + n 2 I 2 =(4A) = 4.86 A (2 A) 46
47 Magnetizing current ripple i M = V 1D T s 2L M To obtain i M = 20% of I M choose L M = V 1D T s 2 i M (28 V)(1 0.35)(5 µs) = 2(4.86 A)(20%) =47µH i M I M 0 v M 0 D T s V 1 i M This leads to a peak magnetizing current (referred to winding 1) of I M,max = I M + i M = 5.83 A 47
48 RMS winding currents Since the winding current ripples are small, the rms values of the winding currents are nearly equal to their dc comonents: I 1 = 4 A I 2 = 2 A Hence the sum of the rms winding currents, referred to the primary, is I tot = I 1 + n 2 I 2 = 4.86 A 48
49 Evaluate K g The following engineering choices are made: Allow 0.75 W of total copper loss (a small core having thermal resistance of less than 40 C/W then would have a temperature rise of less than 30 C) Operate the core at B max = 0.25 T (which is less than the ferrite saturation flux density of 0.3 ot 0.5 T) Use fill factor K u = 0.4 (a reasonable estimate for a lowvoltage inductor with multiple windings) Evaluate K g : K g ( Ω cm)(47 µh) 2 (4.86 A) 2 (5.83 A) 2 (0.25 T) 2 (0.75 W)(0.4) = cm
50 Select core It is decided to use a ferrite PQ core. From Appendix D, the smallest PQ core having K g cm 5 is the PQ 20/16, with K g = cm 5. The data for this core are: A 1 A c = 0.62 cm 2 W A = cm 2 MLT = 4.4 cm 2D 50
51 Air gap length l g = µ 2 0L M I M,max A c B max = (4π 10 7 H/m)(47 µh)(5.83 A) 2 (0.25 T) 2 (0.62 cm 2 ) = 0.52 mm
52 Turns = L MI M,max 10 B max A 4 c (47 µh)(5.83 A) = (0.25 T)(0.62 cm 2 ) 104 = 17.6 turns n 2 = n 2 = (17.6) = 7.54 turns Let s round off to = 17 n 2 = 7 52
53 Wire sizes Allocation of window area: α 1 = I 1 I tot = α 2 = n 2I 2 I tot = (17)(4 A) (17)(4.86 A) = (7)(2 A) (17)(4.86 A) = Determination of wire areas and AWG (from table at end of Appendix D): A w1 α 1K u W A = (0.8235)(0.4)(0.256 cm2 ) = cm 2 (17) use AWG #21 A w2 α 2K u W A = (0.1695)(0.4)(0.256 cm2 ) = cm 2 n 2 (7) use AWG #24 53
54 Example 2: CCM flyback transformer i M Transformer model I M i M V g i 1 i M + L M + v M : n 2 D 1 i 2 C R + V 0 i 1 I M 0 i 2 Q 1 n 2 I M 0 v M V g 0 DT s 54
55 Specifications Input voltage V g = 200V Output (full load) 20 V at 5 A Switching frequency 150 khz Magnetizing current ripple 20% of dc magnetizing current Duty cycle D = 0.4 Turns ratio n 2 / = 0.15 Copper loss 1.5 W Fill factor K u = 0.3 Maximum flux density B max = 0.25 T 55
56 Basic converter calculations Components of magnetizing current, referred to primary: I M = n 2 1 D V R = 1.25 A RMS winding currents: I 1 = I M D i M I M 2 = A i M = 20% I M = 0.25 A I 2 = n 2 I M D i M I M 2 = 6.50 A I M,max = I M + i M = 1.5 A Choose magnetizing inductance: I tot = I 1 + n 2 I 2 = 1.77 A L M = V g DT s 2 i M = 1.07 mh 56
57 Choose core size K g ρl 2 MI 2 2 tot I M,max P cu K u B max = Ωcm H A A 2 = cm T W The smallest EE core that satisfies this inequality (Appendix D) is the EE30. A 57
58 Choose air gap and turns l g = µ 2 0L M I M,max A c B max = 4π 10 7 H/m H 1.5 A T cm =0.44mm = L MI M,max B max A c 10 4 = H 1.5 A 0.25 T 1.09 cm = 58.7 turns Round to =59 58 n 2 = n 2 = = 8.81 n 2 =9
59 Wire gauges α 1 = I 1 I tot = A 1.77 A = 0.45 α 2 = n 2I 2 I tot = A A = 0.55 A W1 α 1K u W A = cm 2 use #28 AWG A W2 α 2K u W A n 2 = cm 2 use #19 AWG 59
60 Core loss CCM flyback example BH loop for this application: B The relevant waveforms: B B sat B max B B max B V g H c Minor B H loop, CCM flyback example 0 v M 0 A c V g DT s B H loop, large excitation B vs. applied voltage, from Faraday s law: db dt = v M A c For the first subinterval: db dt = V g A c 60
61 Calculation of ac flux density and core loss Solve for B: B = V g A c DT s Plug in values for flyback example: 200 V µs B = cm 2 =0.041 T From manufacturer s plot of core loss (at left), the power loss density is 0.04 W/cm 3. Hence core loss is P fe = 0.04 W/cm 3 A c l m = 0.04 W/cm cm cm = 0.25 W 61 Power loss density, Watts/cm W/cm 3 400kHz 200kHz 150kHz 100kHz 50kHz 20kHz B, Tesla
62 Comparison of core and copper loss Copper loss is 1.5 W does not include proximity losses, which could substantially increase total copper loss Core loss is 0.25 W Core loss is small because ripple and B are small It is not a bad approximation to ignore core losses for ferrite in CCM filter inductors Could consider use of a less expensive core material having higher core loss Neglecting core loss is a reasonable approximation for this application Design is dominated by copper loss The dominant constraint on flux density is saturation of the core, rather than core loss 62
63 14.5 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. 63
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