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 Example 2 Multiple-Output Full-Bridge Buck Converter Q 1 D 1 Q 3 D 3 + T 1 : : n 2 D 5 i 2a I 5V 100 A + V g 160 V + i 1 v 1 D 6 i 2b 5 V D Q2 2 D Q4 4 Switching frequency 150 khz Transformer frequency 75 khz Turns ratio 110:5:15 : n 2 : n 3 : n 3 D 7 D 8 i 3a i 2b I 15V 15 A + 15 V Optimize transformer at D =

3 Other transformer design details Use Magnetics, Inc. ferrite P material. Loss parameters at 75 khz: K fe = 7.6 W/T cm 3 = 2.6 Use E-E core shape Assume fill factor of K u = 0.25 (reduced fill factor accounts for added insulation required in multiple-output off-line application) Allow transformer total power loss of P tot = 4 W Use copper wire, with = cm (approximately 0.5% of total output power) 26

4 Applied transformer waveforms D 3 i 1 D 4 T 1 : : n 2 + v 1 : n 2 : n 3 D 5 D 6 D 7 i 2a i 2b i 3a v 1 i 1 i 2a V g n 2 I 5V + n 3 I 15V Area λ 1 = V g DT s 0 n 2 I 5V + n 3 I 15V I 5V 0 0 V g 0.5I 5V 0 : n 3 D 8 i 2b i 3a I 15V 0.5I 15V 0 DT s T s T s +DT s 2T s 0 t 27

5 Applied primary volt-seconds v 1 V g Area λ 1 = V g DT s 0 0 V g λ 1 = DT s V g = (0.75) (6.67 µsec ) (160 V) = 800 V µsec 28

6 Applied primary rms current i 1 n 2 I 5V + n 3 I 15V 0 n 2 I 5V + n 3 I 15V I 1 = n 2 I 5V + n 3 I 15V D = 5.7 A 29

7 Applied rms current, secondary windings i 2a i 3a I 5V I 15V 0.5I 5V 0.5I 15V 0 DT s T s T s +DT s 2T s 0 0 t I 2 = 1 2 I 5V I 3 = 1 2 I 15V 1+D = 66.1 A 1+D = 9.9 A 30

8 I tot RMS currents, summed over all windings and referred to primary I tot = Σall 5 windings n j I j = 5.7 A = 14.4 A = I 1 +2 n 2 I 2 +2 n 3 I A A 31

9 Select core size K gfe ( )( ) 2 (14.4) 2 (7.6) 2/2.6 4 (0.25) (4) 4.6/ = From Appendix D A 32

10 Evaluate ac flux density B Eq. (15.20): B max = 10 8 ρλ I tot 2K u (MLT) W A A c 3 l m 1 βk fe 1 β +2 Plug in values: B = 10 8 ( )( ) 2 (14.4) 2 2(0.25) (8.5) (1.1)(1.27) 3 (7.7) 1 (2.6)(7.6) 1/4.6 =0.23 Tesla This is less than the saturation flux density of approximately 0.35 T 33

11 Evaluate turns Choose according to Eq. (15.21): = λ 1 2 BA c 10 4 = 10 4 ( ) 2(0.23)(1.27) = 13.7 turns Choose secondary turns according to desired turns ratios: n 2 = = 0.62 turns n 3 = = 1.87 turns Rounding the number of turns To obtain desired turns ratio of 110:5:15 we might round the actual turns to 22:1:3 Increased would lead to Less core loss More copper loss Increased total loss 34

12 Loss calculation with rounded turns With = 22, the flux density will be reduced to B = ( ) 2(22)(1.27) 104 = Tesla The resulting losses will be P fe = (7.6)(0.143) 2.6 (1.27)(7.7) = 0.47 W P cu = ( )( ) 2 (14.4) 2 4 (0.25) = 5.4 W P tot = P fe + P cu = 5.9 W (8.5) (1.1)(1.27) 2 1 (0.143) Which exceeds design goal of 4 W by 50%. So use next larger core size: EE50. 35

13 Calculations with EE50 Repeat previous calculations for EE50 core size. Results: B = 0.14 T, = 12, P tot = 2.3 W Again round to 22. Then B = 0.08 T, P cu = 3.89 W, P fe = 0.23 W, P tot = 4.12 W Which is close enough to 4 W. 36

14 Wire sizes for EE50 design Window allocations Wire gauges α 1 = I 1 I tot = = α 2 = n 2I 2 I tot = = α 3 = n 3I 3 I tot = = A w1 = α 1K u W A = (0.396)(0.25)(1.78) = cm (22) 2 AWG #19 A w2 = α 2K u W A = (0.209)(0.25)(1.78) = cm n 2 (1) 2 AWG #8 A w3 = α 3K u W A = (0.094)(0.25)(1.78) = cm n 3 (3) 2 AWG #16 Might actually use foil or Litz wire for secondary windings 37

15 Discussion: Transformer design Process is iterative because of round-off of physical number of turns and, to a lesser extent, other quantities Effect of proximity loss Not included in design process yet Requires additional iterations Can modify procedure as follows: After a design has been calculated, determine number of layers in each winding and then compute proximity loss Alter effective resistivity of wire to compensate: define eff = P cu /P dc where P cu is the total copper loss (including proximity effects) and P dc is the copper loss predicted by the dc resistance. Apply transformer design procedure using this effective wire resistivity, and compute proximity loss in the resulting design. Further iterations may be necessary if the specifications are not met. 38

16 Transformer design

17 Transformer design results Core: ETD 34, TSC 50ALL material

18 15.4 AC Inductor Design i + v L Window area W A Core mean length per turn (MLT) n turns Core Core area A c Air gap l g v Area λ Wire resistivity ρ Fill factor K u t 1 t 2 t i Design a single-winding inductor, having an air gap, accounting for core loss (note that the previous design procedure of this chapter did not employ an air gap, and inductance was not a specification) 39

19 Outline of key equations Obtain specified inductance: L = µ 0A c n 2 l g Relationship between applied volt-seconds and peak ac flux density: B = λ 2nA c Copper loss (using dc resistance): P cu = ρn2 (MLT) K u W A I 2 Total loss is minimized when B = ρλ2 I 2 2K u (MLT) W A A c 3 l m 1 βk fe Must select core that satisfies ρλ 2 I 2 2/β K fe K gfe 2K u P tot β +2/β 1 β +2 See Sectio5.4.2 for step-by-step design equations 40