15.1 Transformer Design: Basic Constraints. Chapter 15: Transformer design. Chapter 15 Transformer Design

Transcription

1 Chapter 5 Transformer Design Some more advanced design issues, not considered in previous chapter: : n Inclusion of core loss Selection of operating flux density to optimize total loss Multiple winding design: as in i (t) v (t) v (t) i (t) the coupled-inductor case, allocate the available window area among several windings R R A transformer design i k (t) procedure How switching frequency v k (t) affects transformer size : n k R k Chapter 5 Transformer Design 5. Transformer design: Basic constraints 5. A step-by-step transformer design procedure 5.3 Examples 5.4 AC inductor design 5.5 Summary 5. Transformer Design: Basic Constraints Core loss P fe = K fe ( B) A c l m Typical value of for ferrite materials:.6 or.7 B is the peak value of the ac component of B(t), i.e., the peak ac flux density So increasing B causes core loss to increase rapidly This is the first constraint 3

2 Flux density Constraint # Flux density B(t) is related to the applied winding voltage according to Faraday s Law. Denote the voltseconds applied to the primary winding during the positive portion of v (t) as : v (t) area λ t t t λ = t t v (t)dt This causes the flux to change from its negative peak to its positive peak. From Faraday s law, the peak value of the ac component of flux density is To attain a given flux density, the primary turns should be chosen according to B = λ n A = c BA c λ 4 Copper loss Constraint #3 Allocate window area between windings in optimum manner, as described in previous section Total copper loss is then equal to with P cu = ρn I tot W A K u I tot = k n j n I j Σj = Eliminate, using result of previous slide: P cu = ρλ I tot 4K u W A A c B Note that copper loss decreases rapidly as B is increased 5 Total power loss 4. P tot = P cu P fe There is a value of B that minimizes the total power loss P tot = P fe P cu Power loss Copper loss P cu P tot Core loss P fe P fe = K fe ( B) A c l m Optimum B B P cu = ρλ I tot 4K u W A A c B 6

3 5. Find optimum flux density B Given that P tot = P fe P cu Then, at the B that minimizes P tot, we can write dp tot d( B) = dp fe d( B) dp cu d( B) =0 Note: optimum does not necessarily occur where P fe = P cu. Rather, it occurs where dp fe d( B) = dp cu d( B) 7 Take derivatives of core and copper loss P fe = K fe ( B) A c l m P cu = ρλ I tot 4K u W A A c B dp fe d( B) = K fe( B) A c l m dp cu I tot d( B) = ρλ 4K u W A A c ( B) 3 Now, substitute into dp fe d( B) = dp cu d( B) and solve for B: B = ρλ I tot K u W A A c 3 l m K fe Optimum B for a given core and application 8 Total loss Substitute optimum B into expressions for P cu and P fe. The total loss is: P tot = A c l m K fe ρλ I tot 4K u W A A c Rearrange as follows: W A A c ( )/ l m / Left side: terms depend on core geometry ρλ I / tot K fe = / 4K u P tot Right side: terms depend on specifications of the application 9

4 The core geometrical constant K gfe ( )/ Define K gfe = W A A c / l m Design procedure: select a core that satisfies ρλ I / tot K fe K gfe / 4K u P tot Appendix D lists the values of K gfe for common ferrite cores K gfe is similar to the K g geometrical constant used in Chapter 4: K g is used when B max is specified K gfe is used when B is to be chosen to minimize total loss 0 5. Step-by-step transformer design procedure Th e following quantities are specified, using the units noted: Wire effective resistivity ( -cm) Total rms winding current, ref to pri I tot (A ) Desired turns ratios n /, n 3 /, etc. Applied pri volt-sec (V -sec) Allowed total power dissipation P tot (W) Winding fill factor K u Core loss exponent Core loss coefficient K fe (W/cm 3 T ) Other quantities and their dimensions: Core cross-sectional area A c (cm ) Core window area W A (cm ) Mean length per turn MLT (cm) Magnetic path length l e (cm) Wire areas A w, (cm ) Peak ac f lux density B (T) Procedure. Determine core size ρλ I / tot K fe K gfe / 08 4K u P tot Select a core from Appendix D that satisfies this inequality. It may be possible to reduce the core size by choosing a core material that has lower loss, i.e., lower K fe.

5 . Evaluate peak ac flux density B = 0 8 ρλ I tot K u W A A c 3 l m K fe At this point, one should check whether the saturation flux density is exceeded. If the core operates with a flux dc bias B dc, then B B dc should be less than the saturation flux density B sat. If the core will saturate, then there are two choices: Specify B using the K g method of Chapter 4, or Choose a core material having greater core loss, then repeat steps and 3 3. and 4. Evaluate turns Primary turns: λ = 0 4 BA c Choose secondary turns according to desired turns ratios: n = n n 3 = n and 6. Choose wire sizes Fraction of window area assigned to each winding: α = I I tot α = n I I tot Choose wire sizes according to: A w α K u W A A w α K u W A n α k = n ki k I tot 5

6 Check: computed transformer model Predicted magnetizing inductance, referred to primary: L M = µa c l m Peak magnetizing current: i (t) i M (t) L M : n i (t) i M, pk = λ L M R R Predicted winding resistances: R = ρ A w R = ρn A w : n k R k i k (t) Example : Single-output isolated Cuk converter I g 4 A v C (t) v C (t) I 0 A V g 5 V i (t) v (t) v (t) n : i (t) V 5 V 00 W f s = 00 khz D = 0.5 n = 5 K u = 0.5 Allow P tot = 0.5 W Use a ferrite pot core, with Magnetics Inc. P material. Loss parameters at 00 khz are K fe = 4.7 =.6 7 Waveforms v (t) V C DT s Area λ D'T s nv C Applied primary voltseconds: λ = DT sv c = (0.5) (5 µsec ) (5 V) = 6.5 Vµsec i (t) I/n Applied primary rms current: I = D n I D' I g =4A i (t) I I g Applied secondary rms current: I = ni =0A ni g Total rms winding current: I tot = I n I =8A 8

7 Choose core size K gfe ( )( ) (8) (4.7) /.6 4 (0.5) (0.5) 4.6/ = Pot core data of Appendix D lists 3 pot core with K gfe = Next smaller pot core is not large enough. 9 Evaluate peak ac flux density B = 0 8 ( )( ) (8) (0.5) = Tesla (4.4) (0.97)(0.635) 3 (3.5) (.6)(4.7) /4.6 This is much less than the saturation flux density of approximately 0.35 T. Values of B in the vicinity of 0. T are typical for ferrite designs that operate at frequencies in the vicinity of 00 khz. 0 Evaluate turns =0 4 ( ) (0.0858)(0.635) = 5.74 turns n = n =.5 turns In practice, we might select = 5 and n = This would lead to a slightly higher flux density and slightly higher loss.

8 Determine wire sizes Fraction of window area allocated to each winding: α = 4A 8A = 0.5 α = Wire areas: 5 0 A = 0.5 8A A w = (0.5)(0.5)(0.97) = cm (5) A w = (0.5)(0.5)(0.97) = cm () (Since, in this example, the ratio of winding rms currents is equal to the turns ratio, equal areas are allocated to each winding) From wire table, Appendix D: AWG #6 AWG #9 Wire sizes: discussion Primary 5 turns #6 AWG Secondary turn #9 AWG Very large conductors! One turn of #9 AWG is not a practical solution Some alternatives Use foil windings Use Litz wire or parallel strands of wire 3 Effect of switching frequency on transformer size for this P-material Cuk converter example Pot core size B max, Tesla khz 50 khz 00 khz 00 khz 50 khz 400 khz 500 khz 000 khz As switching frequency is increased from 5 khz to 50 khz, core size is dramatically reduced Switching frequency As switching frequency is increased from 400 khz to MHz, core size increases 0 4

9 5.3. Example Multiple-Output Full-Bridge Buck Converter Q D Q 3 D 3 T : : n D 5 i a (t) I 5V 00 A V g 60 V i (t) v (t) D 6 i b (t) 5 V D D 4 Q Q4 Switching frequency 50 khz Transformer frequency 75 khz Turns ratio 0:5:5 : n : n 3 D 7 D 8 : n 3 i 3a (t) i b (t) I 5V 5 A 5 V Optimize transformer at D = Other transformer design details Use Magnetics, Inc. ferrite P material. Loss parameters at 75 khz: K fe = 7.6 W/T cm 3 =.6 Use E-E core shape Assume fill factor of K u = 0.5 (reduced fill factor accounts for added insulation required in multiple-output off-line application) Allow transformer total power loss of P tot = 4 W (approximately 0.5% of total output power) Use copper wire, with = cm 6 Applied transformer waveforms T D 3 : : n D 5 i (t) v (t) D 6 : n D 4 : n 3 i a (t) i b (t) i 3a (t) v (t) i (t) V g n I 5V n 3 I 5V 0 n I 5V n 3 I 5V Area λ = V g DT s 0 0 V g D 7 i a (t) I 5V 0.5I 5V 0 : n 3 D 8 i b (t) i 3a (t) I 5V 0.5I 5V 0 DT s T s T s DT s T s 0 t 7

10 Applied primary volt-seconds v (t) V g Area λ = V g DT s 0 0 V g λ = DT s V g = (0.75) (6.67 µsec ) (60 V) = 800 Vµsec 8 Applied primary rms current i (t) n I 5V n 3 I 5V 0 n I 5V n 3 I 5V I = n I 5V n 3 I 5V D = 5.7 A 9 Applied rms current, secondary windings i a (t) i 3a (t) I 5V I 5V 0.5I 5V 0.5I 5V 0 DT s T s T s DT s T s 0 0 t I = I 5V I 3 = I 5V D = 66. A D = 9.9 A 30

11 I tot RMS currents, summed over all windings and referred to primary I tot = Σall 5 windings n j I j = 5.7 A 0 5 = 4.4 A = I n I n 3 I A A 3 Select core size K gfe ( )( ) (4.4) (7.6) /.6 4 (0.5) (4) 4.6/ = From Appendix D A 3 Evaluate ac flux density B Eq. (5.0): Plug in values: B max = 0 8 ρλ I tot K u W A A c 3 l m K fe B = 0 8 ( )( ) (4.4) (0.5) =0.3 Tesla (8.5) (.)(.7) 3 (7.7) (.6)(7.6) /4.6 This is less than the saturation flux density of approximately 0.35 T 33

12 Evaluate turns Choose according to Eq. (5.): λ = 0 4 BA c =0 4 ( ) (0.3)(.7) = 3.7 turns Choose secondary turns according to desired turns ratios: n = 5 0 = 0.6 turns n 3 = 5 0 =.87 turns Rounding the number of turns To obtain desired turns ratio of 0:5:5 we might round the actual turns to ::3 Increased would lead to Less core loss More copper loss Increased total loss 34 Loss calculation with rounded turns With =, the flux density will be reduced to B = ( ) ()(.7) 04 = 0.43 Tesla The resulting losses will be P fe = (7.6)(0.43).6 (.7)(7.7) = 0.47 W P cu = ( )( ) (4.4) 4 (0.5) = 5.4 W P tot = P fe P cu = 5.9 W (8.5) (.)(.7) (0.43) 08 Which exceeds design goal of 4 W by 50%. So use next larger core size: EE Calculations with EE50 Repeat previous calculations for EE50 core size. Results: B = 0.4 T, =, P tot =.3 W Again round to. Then B = 0.08 T, P cu = 3.89 W, P fe = 0.3 W, P tot = 4. W Which is close enough to 4 W. 36

13 Wire sizes for EE50 design Window allocations Wire gauges α = I I tot = = α = n I I tot = = 0.09 α 3 = n 3I 3 I tot = = A w = α K u W A = (0.396)(0.5)(.78) = cm () AWG #9 A w = α K u W A = (0.09)(0.5)(.78) = cm n () AWG #8 A w3 = α 3K u W A = (0.094)(0.5)(.78) = cm n 3 (3) AWG #6 Might actually use foil or Litz wire for secondary windings 37 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 AC Inductor Design v(t) i(t) v(t) L Area λ Window area W A Core mean length per turn n turns Wire resistivity ρ Fill factor K u Core Core area A c Air gap l g t t t i(t) 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

14 Outline of key equations Obtain specified inductance: L = µ 0A c n l g Relationship between applied volt-seconds and peak ac flux density: B = na λ c Copper loss (using dc resistance): P cu = ρn I K u W A Total loss is minimized when B = ρλ I K u W A A c 3 l m K fe Must select core that satisfies ρλ I / K fe K gfe / K u P tot See Sectio5.4. for step-by-step design equations 40