R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

Transcription

1 R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

2 Low-frequency copper loss DC resistance of wire R = ρ l b A w where A w is the wire bare cross-sectional area, and l b is the length of the wire. The resistivity is equal to cm for soft-annealed copper at room temperature. This resistivity increases to cm at 100 C. i(t) R The wire resistance leads to a power loss of 2 P cu = I rms R 3

3 13. Eddy currents in winding conductors Intro to the skin and proximity effects Wire Φ(t) Current density i(t) Eddy currents Wire δ i(t) Eddy currents

4 Penetration depth For sinusoidal currents: current density is an exponentially decaying function of distance into the conductor, with characteristic length known as the penetration depth or skin depth. Penetration depth δ, cm 0.1 Wire diameter #20 AWG δ = ρ πµ f C 25 C #30 AWG #0 AWG For copper at room temperature: δ = 7.5 f cm khz 100 khz 1 MHz Frequency 5

5 The proximity effect Ac current in a conductor induces eddy currents in adjacent conductors by a process called the proximity effect. This causes significant power loss in the windings of high-frequency transformers and ac inductors. A multi-layer foil winding, with h >. Each layer carries net current i(t). Current density J iconductor 1 h Area Φ Conductor 2 i i i Area i Area i 6

6 Example: a two-winding transformer Cross-sectional view of two-winding transformer example. Primary turns are wound in three layers. For this example, let s assume that each layer is one turn of a flat foil conductor. The secondary is a similar three-layer winding. Each layer carries net current i(t). Portions of the windings that lie outside of the core window are not illustrated. Each layer has thickness h >. Core i i i { Layer 1 Layer 2 Layer 3 Primary winding i i i Layer 3 { Layer 2 Layer 1 Secondary winding 7

7 Distribution of currents on surfaces of conductors: two-winding example Skin effect causes currents to concentrate on surfaces of conductors Surface current induces equal and opposite current on adjacent conductor This induced current returns on opposite side of conductor Net conductor current is equal to i(t) for each layer, since layers are connected in series Circulating currents within layers increase with the numbers of layers Core Current density J Layer 1 h Layer 1 i i i Layer 2 Primary winding 8 Layer 3 i i i Layer 3 Layer 2 Φ 2Φ 3Φ 2Φ Φ i i 2i 2i 3i 3i 2i 2i i i Layer 2 Layer 3 Layer 3 Layer 2 Layer 1 Layer 1 Secondary winding

8 Estimating proximity loss: high-frequency limit The current i(t) having rms value I is confined to thickness d on the surface of layer 1. Hence the effective ac resistance of layer 1 is: R ac = h δ R dc This induces copper loss P 1 in layer 1: P 1 = I 2 R ac Power loss P 2 in layer 2 is: P 2 = P 1 +P 1 =5P 1 Power loss P 3 in layer 3 is: Current density J h Layer 1 Φ 2Φ 3Φ 2Φ Φ i i 2i 2i 3i 3i 2i 2i i i Layer 2 Primary winding Layer 3 Layer 3 Layer 2 Secondary winding Power loss P m in layer m is: Layer 1 P 3 = P 1 = 13P 1 P m = I 2 m m 2 h δ R dc 9

9 Total loss in M-layer winding: high-frequency limit Add up losses in each layer: Μ Σ P = I 2 h δ R dc m m 2 m =1 = I 2 h δ R M dc 2M Compare with dc copper loss: If foil thickness were H =, then at dc each layer would produce copper loss P 1. The copper loss of the M-layer winding would be P dc = I 2 MR dc So the proximity effect increases the copper loss by a factor of F R = P P dc = 1 3 h δ 2M

10 13..2 Leakage flux in windings A simple two-winding transformer example: core and winding geometry Each turn carries net current i(t) in direction shown Primary winding { Secondary winding { y x Core µ > µ 0 51

11 Flux distribution Mutual flux M is large and is mostly confined to the core Leakage flux is present, which does not completely link both windings Because of symmetry of winding geometry, leakage flux runs approximately vertically through the windings Leakage flux Mutual flux Φ M 52

12 Analysis of leakage flux using Ampere s law Ampere s law, for the closed path taken by the leakage flux line illustrated: Leakage path Enclosed current = F(x)=H(x)l w (note that MMF around core is small compared to MMF through the air inside the winding, because of high permeability of core) Enclosed current H(x) + F(x) l w x 53

13 Ampere s law for the transformer example For the innermost leakage path, enclosing the first layer of the primary: This path encloses four turns, so the total enclosed current is i(t). For the next leakage path, enclosing both layers of the primary: This path encloses eight turns, so the total enclosed current is 8i(t). The next leakage path encloses the primary plus four turns of the secondary. The total enclosed current is 8i(t) i(t) = i(t). Leakage flux Mutual flux Φ M 5

14 MMF diagram, transformer example MMF F(x) across the core window, as a function of position x Enclosed current = F(x)=H(x)l w Leakage path Primary winding { Layer 1 Layer 2 Secondary winding { Layer 2 Layer 1 Enclosed current F(x) l w F(x) 8i H(x) + i x 0 x 55

15 Two-winding transformer example Winding layout Core i i i i i i Layer 1 Layer 2 Layer 3 Layer 3 Layer 2 Layer 1 MMF diagram Use Ampere s law around a closed path taken by a leakage flux line: m p m s i = F(x) m p = number of primary layers enclosed by path m s = number of secondary layers enclosed by path MMF F(x) 3i 2i i 0 i i i i i i x 56

16 Two-winding transformer example with proximity effect Flux does not penetrate conductors Surface currents cause net current enclosed by leakage path to be zero when path runs down interior of a conductor Magnetic field strength H(x) within the winding is given by H(x)= F(x) l w MMF F(x) 3i 2i i 0 i i 2i 2i 3i 3i 2i 2i i i x 57

17 Interleaving the windings: MMF diagram pri sec pri sec pri sec MMF F(x) i i i i i i 3i 2i i 0 x Greatly reduces the peak MMF, leakage flux, and proximity losses 58

18 A partially-interleaved transformer For this example, there are three primary layers and four secondary layers. The MMF diagram contains fractional values. MMF F(x) 1.5 i i 0.5 i i i 1.5 i Secondary 3i m = 1 3i m = 2 Primary i i i 3i m = 1.5 m = 0.5 m = 1.5 Secondary m = 2 3i m = 1 x 59

19 13..3 Foil windings and layers d Approximating a layer of round conductors as an effective single foil conductor: (a) (b) (c) (d) h h h l w Square conductors (b) have same crosssectional area as round conductors (a) if h = π d Eliminate space between square conductors: push together into a single foil turn (c) (d) Stretch foil so its width is l w. The adjust conductivity so its dc resistance is unchanged 60

20 Winding porosity Stretching the conductor increases its area. Compensate by increasing the effective resistivity, to maintain the same dc resistance. Define winding porosity as the ratio of cross-sectional areas. If layer of width l w contains n l turns of round wire having diameter d, then the porosity is (c) h (d) h η = π d n l l w Typical for full-width round conductors is = 0.8. The increased effective resistivity increases the effective skin depth: δ = δ η l w Define = h/d. The effective value for a layer of round conductors is ϕ = h δ = η π d δ 61

21 13.. Power loss in a layer Approximate computation of copper loss in one layer Assume uniform magnetic fields at surfaces of layer, of strengths H(0) and H(h). Assume that these fields are parallel to layer surface (i.e., neglect fringing and assume field normal component is zero). H(0) Layer H(h) The magnetic fields H(0) and H(h) are driven by the MMFs F(0) and F(h). Sinusoidal waveforms are assumed, and rms values are used. It is assumed that H(0) and H(h) are in phase. F(x) F(0) F(h) 0 h 62

22 Solution for layer copper loss P Solve Maxwell s equations to find current density distribution within layer. Then integrate to find total copper loss P in layer. Result is P = R dc ϕ n l 2 F 2 (h)+f 2 (0) G 1 (ϕ) F(h)F(0)G 2 (ϕ) where R dc = ρ l b = ρ (MLT)n 3 l A w G 1 (ϕ)= ηl w 2 sinh(2ϕ) +sin (2ϕ) cosh (2ϕ) cos (2ϕ) n l = number of turns in layer, R dc = dc resistance of layer, (MLT) = mean-length-per-turn, or circumference, of layer. G 2 (ϕ)= sinh (ϕ) cos (ϕ)+cosh (ϕ) sin (ϕ) cosh (2ϕ) cos (2ϕ) ϕ = h δ = η π d δ η = π d n l l w 63

23 Winding carrying current I, with n l turns per layer If winding carries current of rms magnitude I, then F(h) F(0) = n l I Express F(h) in terms of the winding current I, as F(h)=mn l I H(0) Layer H(h) The quantity m is the ratio of the MMF F(h) to the layer ampere-turns n l I. Then, F(0) F(h) = m m 1 Power dissipated in the layer can now be written P = I 2 R dc ϕq (ϕ, m) F(x) F(0) F(h) Q (ϕ, m)= 2m 2 2m +1 G 1 (ϕ) mm 1 G 2 (ϕ) 0 h 6

24 Increased copper loss in layer P = I 2 R dc ϕq (ϕ, m) 100 m = P I 2 R dc m = ϕ 65

25 Layer copper loss vs. layer thickness P P dc ϕ =1 = Q (ϕ, m) P P dc ϕ =1 100 m = Relative to copper loss when h = m = ϕ 66

26 13..5 Example: Power loss in a transformer winding Two winding transformer Each winding consists of M layers Proximity effect increases copper loss in layer m by the factor Q (,m) Sum losses over all primary layers: F R = P pri P pri,dc = 1 M M Σ m =1 F Mn p i 2n p i n p i 0 ϕq (ϕ, m) Primary layers Secondary layers { { n p i n p i n p i n p i n p i n p i m = 1 m = 2 m = M m = M m = 2 m = 1 x 67

27 Express summation in closed form: Increased total winding loss F R = ϕ G 1 (ϕ)+ 2 3 M 2 1 G 1 (ϕ) 2G 2 (ϕ) 100 Number of layers M = F R = P pri P pri,dc ϕ 68

28 Total winding loss P pri P pri,dc ϕ =1 = G 1 (ϕ)+ 2 3 M 2 1 G 1 (ϕ) 2G 2 (ϕ) P pri P pri,dc ϕ = Number of layers M = ϕ 69

29 13..6 Interleaving the windings Same transformer example, but with primary and secondary layers alternated Each layer operates with F = 0 on one side, and F = i on the other side Proximity loss of entire winding follows M = 1 curve MMF F(x) 3i 2i i 0 pri sec pri sec pri sec i i i i i i For M = 1: minimum loss occurs with = /2, although copper loss is nearly constant for any 1, and is approximately equal to the dc copper loss obtained when = 1. x 70

30 Partial interleaving Partially-interleaved example with 3 primary and secondary layers Each primary layer carries current i while each secondary layer carries 0.75i. Total winding currents add to zero. Peak MMF occurs in space between windings, but has value 1.5i. MMF F(x) 1.5 i i 0.5 i i i 1.5 i Secondary 3i m = 1 3i m = 2 Primary i i i 3i m = 1.5 m = 0.5 m = 1.5 Secondary m = 2 3i m = 1 x We can apply the previous solution for the copper loss in each layer, and add the results to find the total winding losses. To determine the value of m to use for a given layer, evaluate F(h) m = F(h) F(0) 71

31 Determination of m m = m = m = m = Leftmost secondary layer: F(h) F(h) F(0) = 0.75i 0.75i 0 =1 Next secondary layer: F(h) F(h) F(0) = 1.5i 1.5i ( 0.75i) = 2 Next layer (primary): F(0) F(0) F(h) = 1.5i 1.5i ( 0.5i) = 1.5 Center layer (primary): MMF F(x) 1.5 i i 0.5 i i i 1.5 i Secondary 3i m = 1 3i m = 2 Primary i i i 3i m = 1.5 m = 0.5 m = 1.5 Secondary F(h) F(h) F(0) = 0.5i 0.5i ( 0.5i) = 0.5 Use the plot for layer loss (repeated on next slide) to find loss for each layer, according to its value of m. Add results to find total loss. m = 2 3i m = 1 x 72

32 Layer copper loss vs. layer thickness P P dc ϕ =1 = Q (ϕ, m) P P dc ϕ =1 100 m = Relative to copper loss when h = m = ϕ 73

33 Discussion: design of winding geometry to minimize proximity loss Interleaving windings can significantly reduce the proximity loss when the winding currents are in phase, such as in the transformers of buckderived converters or other converters In some converters (such as flyback or SEPIC) the winding currents are out of phase. Interleaving then does little to reduce the peak MMF and proximity loss. See Vandelac and Ziogas [10]. For sinusoidal winding currents, there is an optimal conductor thickness near = 1 that minimizes copper loss. Minimize the number of layers. Use a core geometry that maximizes the width l w of windings. Minimize the amount of copper in vicinity of high MMF portions of the windings 7

34 Litz wire A way to increase conductor area while maintaining low proximity losses Many strands of small-gauge wire are bundled together and are externally connected in parallel Strands are twisted, or transposed, so that each strand passes equally through each position on inside and outside of bundle. This prevents circulation of currents between strands. Strand diameter should be sufficiently smaller than skin depth The Litz wire bundle itself is composed of multiple layers Advantage: when properly sized, can significantly reduce proximity loss Disadvantage: increased cost and decreased amount of copper within core window 75

35 13..7 PWM waveform harmonics Fourier series: with i(t)=i 0 + Σ j =1 2 I j cos (jωt) i(t) I pk I j = 2 I pk jπ sin (jπd) I 0 = DI pk Copper loss: 0 DT s T s t Dc P dc = I 0 2 R dc Ac P j = I j 2 R dc j ϕ 1 G 1 ( j ϕ 1 )+ 2 3 M 2 1 G 1 ( j ϕ 1 ) 2G 2 ( j ϕ 1 ) Total, relative to value predicted by low-frequency analysis: P cu DI pk 2 = D + 2ϕ 1 R dc Dπ 2 Σ j =1 sin 2 (jπd) j j G 1 ( j ϕ 1 )+ 2 3 M 2 1 G 1 ( j ϕ 1 ) 2G 2 ( j ϕ 1 ) 76

36 Harmonic loss factor F H Effect of harmonics: F H = ratio of total ac copper loss to fundamental copper loss Σ P j j =1 F H = P 1 The total winding copper loss can then be written P cu = I 0 2 R dc + F H F R I 1 2 R dc 77

37 Increased proximity losses induced by PWM waveform harmonics: D = D = 0.5 M = 10 F H M = ϕ

38 Increased proximity losses induced by PWM waveform harmonics: D = 0.3 F H 100 D = M = M = ϕ 1 79

39 Increased proximity losses induced by PWM waveform harmonics: D = 0.1 F H M = 10 D = M = ϕ 1 80

40 Discussion: waveform harmonics Harmonic factor F H accounts for effects of harmonics Harmonics are most significant for 1 in the vicinity of 1 Harmonics can radically alter the conclusion regarding optimal wire gauge A substantial dc component can drive the design towards larger wire gauge Harmonics can increase proximity losses by orders of magnitude, when there are many layers and when 1 lies in the vicinity of 1 For sufficiently small 1, F H tends to the value 1 + (THD) 2, where the total harmonic distortion of the current is THD = Σ I j 2 j =2 I 1 81