Chapter 14 Inductor Design 14.1 Filter inductor design constraints 14.2 A step-by-step design procedure 14.3 Multiple-winding magnetics design using the K g method 14.4 Examples 14.5 Summary of key points 1
14.1 Filter inductor design constraints Objective: i L R Design inductor having a given inductance L, which carries worst-case current I max without saturating, and which has a given winding resistance R, or, equivalently, exhibits a worst-case 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
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
14.1.1 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 worst-case 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
14.1.2 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
14.1.3 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
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 low-voltage inductor 0.25 to 0.3 for off-line transformer 0.05 to 0.2 for high-voltage transformer (multiple kv) 0.65 for low-voltage foil-winding inductor 7
14.1.4 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 1.724 10 6 -cm. The length of the wire comprising an n-turn winding can be expressed as l b = n (MLT) where (MLT) is the mean-length-per-turn of the winding. The meanlength-per-turn 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
14.1.5 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
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 Right-hand side: specifications or other known quantities Left-hand 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
Discussion K g = A 2 cw A (MLT) ρl2 2 I max 2 B max RK u K g is a figure-of-merit 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
14.2 A step-by-step 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 cross-sectional 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
Determine core size K g ρl2 2 I max 2 10 8 (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
Determine air gap length l g = µ 2 0LI max 2 10 4 A c B max (m) with A c expressed in cm 2. µ 0 = 4 10 7 H/m. The air gap length is given in meters. The value expressed above is approximate, and neglects fringing flux and other nonidealities. 14
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 2 10 9 A c 2 (Henries) Units: A c cm 2, L Henries, B max Tesla. 15
Determine number of turns n n = LI max B max A c 10 4 16
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
14.3 Multiple-winding 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 step-by-step design procedure 18
14.3.1 Window area allocation Given: application with k windings having known rms currents and desired turns ratios v 1 n 1 = v 2 n 2 = = v k n k rms current I 1 n 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
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 2 + + α k =1 20
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
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
Variation of copper losses with 1 Copper loss Pcu,1 P cu,tot P cu,2 + P cu,3 +...+ 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
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 2 + + 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
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
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
Example PWM full-bridge transformer i 1 i 2 I n 1 turns { } n 2 turns } n 2 turns i 1 n 2 n 1 I i 3 0 0 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 n 1 I 0.5I 0.5I 0 I 0.5I 0.5I 0 DT s T s T s +DT s 2T s t 27
Expressions for RMS winding currents i 1 n 2 n 1 I I 1 = 1 2Ts I 2 = I 3 = 1 2Ts see Appendix A 0 2T s i 2 1 dt = n 2 I n 1 0 D 2T s i 2 2 dt = 1 2 I 1+D i 2 i 3 I 0 0 0 n 2 n 1 I 0.5I 0.5I 0 I 0.5I 0.5I 0 DT s T s T s +DT s 2T s t 28
Allocation of window area: α m = V mi m Σ V j I j n =1 Plug in rms current expressions. Result: α 1 = 1 1+ 1+D D Fraction of window area allocated to primary winding α 2 = α 3 = 1 2 1 1+ D 1+D Fraction of window area allocated to each secondary winding 29
Numerical example Suppose that we decide to optimize the transformer design at the worst-case operating point D = 0.75. Then we obtain α 1 = 0.396 α 2 = 0.302 α 3 = 0.302 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
14.3.2 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 1 : n 2 + v 2 i 2 Magnetic circuit model: R c R 1 R 2 n 1 i M + Φ R g + v k i k : n k R k 31
14.4 Examples 14.4.1 Coupled Inductor for a Two-Output Forward Converter 14.4.2 CCM Flyback Transformer 32
14.4.2 Example 2: CCM flyback transformer i M Transformer model I M i M V g i 1 i M + L M + v M n 1 : n 2 D 1 i 2 C R + V 0 i 1 I M 0 i 2 Q 1 n 1 n 2 I M 0 v M V g 0 DT s 33
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 /n 1 = 0.15 Copper loss 1.5 W Fill factor K u = 0.3 Maximum flux density B max = 0.25 T 34
Basic converter calculations Components of magnetizing current, referred to primary: I M = n 2 n 1 1 D V R = 1.25 A RMS winding currents: I 1 = I M D 1+ 1 3 i M I M 2 = 0.796 A i M = 20% I M = 0.25 A I 2 = n 1 n 2 I M D 1+ 1 3 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 n 1 I 2 = 1.77 A L M = V g DT s 2 i M = 1.07 mh 35
Choose core size K g ρl 2 MI 2 2 tot I M,max 2 10 8 P cu K u B max = 1.724 10 6 Ω-cm 1.07 10 3 H 2 1.77 A 2 1.5 A 2 = 0.049 cm 5 0.25 T 2 1.5 W 0.3 10 8 The smallest EE core that satisfies this inequality (Appendix D) is the EE30. 36
Choose air gap and turns l g = µ 2 0L M I M,max 2 10 4 A c B max = 4π 10 7 H/m 1.07 10 3 H 1.5 A 2 0.25 T 2 1.09 cm 2 10 4 =0.44mm n 1 = L MI M,max B max A c 10 4 = 1.07 10 3 H 1.5 A 0.25 T 1.09 cm 2 10 4 = 58.7 turns Round to n 1 =59 37 n 2 = n 2 n 1 n 1 = 0.15 59 = 8.81 n 2 =9
Wire gauges α 1 = I 1 I tot = 0.796 A 1.77 A = 0.45 α 2 = n 2I 2 n 1 I tot = 9 6.5 A 59 1.77 A = 0.55 A W1 α 1K u W A n 1 = 1.09 10 3 cm 2 use #28 AWG A W2 α 2K u W A n 2 = 8.88 10 3 cm 2 use #19 AWG 38
Core loss CCM flyback example B-H 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 n 1 A c V g DT s B H loop, large excitation B vs. applied voltage, from Faraday s law: db dt = v M n 1 A c For the first subinterval: db dt = V g n 1 A c 39
Calculation of ac flux density and core loss Solve for B: B = V g n 1 A c DT s Plug in values for flyback example: 200 V 0.4 6.67 µs B = 10 4 259 1.09 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 3 1.09 cm 2 5.77 cm = 0.25 W 40 Power loss density, Watts/cm 3 1 0.1 0.04 W/cm 3 400kHz 200kHz 150kHz 100kHz 50kHz 20kHz 0.01 0.01 0.1 0.3 0.041 B, Tesla
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 41
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 low-frequency 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. 42