SlideShare a Scribd company logo

Inductor design

•
•
B
binodsahu8

The document describes the design of a 250V, 2.565A single-phase AC inductor. It involves calculating the required inductance, selecting a suitable core, determining the number of turns, calculating losses, and ensuring design constraints are met. The key steps are: 1) Calculating inductance and selecting a core that meets the area product requirement 2) Determining the number of turns needed to achieve the required inductance 3) Ensuring operating flux density and losses meet specifications by calculating copper, core, and gap losses.

1 of 9
Downloaded 62 times
1 Design Experiment # 2 Design of a single-phase AC Inductor Related theory: - Design of an AC inductor is quite similar to that of a transformer. The main difference is it consists of only one winding and an air-gap (for designing linear inductor). Design of inductor requires calculation of power handling capability. In some applications the required inductance is specified and in other current rating is specified. If inductance is specified, current rating can be calculated or if current rating is specified, value of inductance can be calculated as follows: If inductance ‘L’ is specified, then inductive reactance fLwLXL 2 in . Then current rating of the inductor is Ampin X V I L  . If current rating is specified then inductive reactance,  in I V XL . Then inductance of the inductor ‘L’ is Heneryin f X L L 2  . Relationship between area product '' pA and power handling capability '' tP of inductor: 4 4 10 cmin KfJKB P A ufac t p   . (1) Where inVAIVPt  .
2 Relationship between core geometry '' gK and power handling capability '' tP of inductor: 5 cmin K P K e t g   . 5 2 cmin MLT KAW K uca g  . Electric and magnetic loading condition, 4222 10145.0   mfe BfkK Number of turns required for inductor design cacf fABk V N 4 10  turns. (2) Inductance of iron-core inductor with air-gap is i N L   (3) S N S Ni i N L 2  Where ‘S’ is the total reluctance of the magnetic circuit. Total reluctance        r i g c gi l l A SSS 0 1 (4) Putting value of ''S in equation (3) we get r i g c l l AN L     2 0 Therefore inductance of the inductor
3 Heneryin MPL l AN L r g c             82 104.0 . (5) Where ''MPL is the magnetic path length of the iron path same as '' il . From equation (5) length of the required air-gap can be expressed as           r c g MPL L AN l   82 104.0 (6) If the core air-gap length '' gl is more as compared to r MPL  then inductance of the inductor becomes Heneryin l AN L g c 82 104.0     (7) Effect of fringing flux: - Final determination of effective air-gap requires the consideration of the effect of ‘fringing flux’ which is a function of gap dimension, shape of pole faces, and shape, size and location of winding. Figure 1 Fringing Flux.
4 Figure 2 Fringing Flux in EI core. The main consequence of the fringing effect is to make the magnetic flux density of the air gap different from the flux density of the core. Fringing in magnetic circuit increases the cross- sectional area of air gap thereby reducing the circuit reluctance (we approximately account for fringing by adding the gap length to the depth and width of iron in computing effective gap area). This increases the inductance of the inductor by a factor ‘F’. Fringing flux factor empirically can be calculated as follows          gc g l G A l F 2 ln1 (8) New value of inductance due to fringing is g c l FAN FLL 82 104.0 '      (9) To maintain the predefined value of ‘L’, as reluctance decreases, number of turns should be reduced        S N L 2 . Now the new value of number of turns required for the inductor is
5 8 104.0    FA Ll N c g new  (10) Now with new number of turns, the operating flux density can be obtained using equation (2) cnewf ac fANk V B 4 10  (11) Design a 250 V, 2.565 A single-phase AC inductor Applied voltage, V = 250 Volts Line current, I = 2.565 A Frequency = 50 Hertz. Current density, J = 300 Amp/cm2 Efficiency goal, 𝜂= 85 % Magnetic material = Silicon Magnetic material permeability, r = 1500 Flux density, Bac = 1.4 Tesla Window utilization, Ku = 0.4 Temperature rise goal, Tr = 50°C Step-1:- Calculation of apparent power VAVIP Lt 25.641565.2250  . Step-2:- Calculation of area product AP 4 44 94.171 3004.1504.044.4 1025.64110 cm JfBKk P A acuf t p       Step-3:- Selection of Core Core material suitable for designing the inductor is EI-150 for which the area product is 150.136 cm4 . Tables blow depicts the design and dimensional data for various laminations EI (Table 3.3 of Text Book).
6 Table 1: Design data for various EI laminations. Table 2: Dimensional data for various EI laminations. E D D Figure 3 Outline of EI laminations. For 𝑬𝑰 − 𝟏𝟓𝟎 lamination, Magnetic path length (MPL) = 22.9 cm Core weight = 2.334 Kg Copper weight = 853 gm
7 Mean length turn (MLT) = 22 cm Iron area, 𝐴 𝑐 = 13.79 cm2 Window area, 𝑊𝑎 = 10.89 cm2 Area product, 𝐴 𝑝 = 𝐴 𝑐 × 𝑊𝑎 = 150.136 cm4 Core geometry, 𝐾𝑔 = 37.579 cm5 Surface area, 𝐴 𝑡 = 479 cm2 Window length, G = 5.715 cm Lamination tongue, E = 3.81 cm Step-4:- Calculation of number of turns 584 79.13504.144.4 1025010 44       cacf fABk V N turns. Step-5: - Calculation of inductive reactance and inductance Inductive reactance,  466.97 565..2 250 L L I V X Inductance, H f X L L 31.0 502 466.97 2     Step-6:- Calculation of required air-gap mmcm MPL L AN l r c g 75.1175.0 1500 9.22 31.0 1079.135844.0104.0 8282                Step-7:- Calculation of Fringing flux factor 197.1 175.0 715.52 ln 79.13 175.0 1 2 ln1                  gc g l G A l F Step-8:- Calculation of new number of turns for the inductor 512 10197.179.134.0 175.031.0 104.0 88         FA Ll N c g new turns. Step-9:- Calculation of operating flux density with new turns 595.1 79.135051244.4 1025010 44       cnewf ac fANk V B T
8 Step-10:- Calculation of bare conductor area 22 )( 855.000855.0 300 565.2 mmcm J I A Bw  Step-11:- Selection of wire from wire table Cross-sectional area of the required wire close to the area calculated in step-10 is matching with the area of SWG-19 conductor. So the selected wire for designing the inductor is SWG-19. Table 3: Standard Wire Gauge table. Standard wire gauge Diameter in mm Cross-sectional area in mm2 Resistance per length in Ω/Km in µΩ/cm 18 1.22 1.17 14.8 148 19 1.02 0.811 21.3 213 20 0.914 0.657 26.3 263 For 19-SWG wire, bare conductor area if 0.811 mm2 and resistance is 213 cm/ . Step-12:- Calculation of inductor resistance Inductor resistance,   4.2102135122210213 66 newL NMLTR Step-13:- Inductor winding copper loss 79.154.2565.2 22  LL RIP Watts Step-14: - Calculation of Watts/Kg       K W 95.0595.150000557.0000557.0 86.168.186.168.1  acBf K W Step-15: - Calculation of core loss 22.2334.295.0  tfefe W K W P Watts Step-16: - Calculation of gap loss 146.13595.150175.081.3155.0 22  acgig fBElKP Watts Step-17: - Calculation of total losses 136.31146.132.279.15  gfecu PPPP Watts
9 Step-18: - Calculation inductor surface watt density 065.0 479 136.31   tA P  Watts/cm2 Step-19: - Calculation of temperature rise CTr 0826.0826.0 5.40065.0400400   Step-20:- Calculation of window utilization factor 38.0 887.10 00811.0512      a wBnew u w AN K

More Related Content

Inductor design

  • 1. 1 Design Experiment # 2 Design of a single-phase AC Inductor Related theory: - Design of an AC inductor is quite similar to that of a transformer. The main difference is it consists of only one winding and an air-gap (for designing linear inductor). Design of inductor requires calculation of power handling capability. In some applications the required inductance is specified and in other current rating is specified. If inductance is specified, current rating can be calculated or if current rating is specified, value of inductance can be calculated as follows: If inductance ‘L’ is specified, then inductive reactance fLwLXL 2 in . Then current rating of the inductor is Ampin X V I L  . If current rating is specified then inductive reactance,  in I V XL . Then inductance of the inductor ‘L’ is Heneryin f X L L 2  . Relationship between area product '' pA and power handling capability '' tP of inductor: 4 4 10 cmin KfJKB P A ufac t p   . (1) Where inVAIVPt  .
  • 2. 2 Relationship between core geometry '' gK and power handling capability '' tP of inductor: 5 cmin K P K e t g   . 5 2 cmin MLT KAW K uca g  . Electric and magnetic loading condition, 4222 10145.0   mfe BfkK Number of turns required for inductor design cacf fABk V N 4 10  turns. (2) Inductance of iron-core inductor with air-gap is i N L   (3) S N S Ni i N L 2  Where ‘S’ is the total reluctance of the magnetic circuit. Total reluctance        r i g c gi l l A SSS 0 1 (4) Putting value of ''S in equation (3) we get r i g c l l AN L     2 0 Therefore inductance of the inductor
  • 3. 3 Heneryin MPL l AN L r g c             82 104.0 . (5) Where ''MPL is the magnetic path length of the iron path same as '' il . From equation (5) length of the required air-gap can be expressed as           r c g MPL L AN l   82 104.0 (6) If the core air-gap length '' gl is more as compared to r MPL  then inductance of the inductor becomes Heneryin l AN L g c 82 104.0     (7) Effect of fringing flux: - Final determination of effective air-gap requires the consideration of the effect of ‘fringing flux’ which is a function of gap dimension, shape of pole faces, and shape, size and location of winding. Figure 1 Fringing Flux.
  • 4. 4 Figure 2 Fringing Flux in EI core. The main consequence of the fringing effect is to make the magnetic flux density of the air gap different from the flux density of the core. Fringing in magnetic circuit increases the cross- sectional area of air gap thereby reducing the circuit reluctance (we approximately account for fringing by adding the gap length to the depth and width of iron in computing effective gap area). This increases the inductance of the inductor by a factor ‘F’. Fringing flux factor empirically can be calculated as follows          gc g l G A l F 2 ln1 (8) New value of inductance due to fringing is g c l FAN FLL 82 104.0 '      (9) To maintain the predefined value of ‘L’, as reluctance decreases, number of turns should be reduced        S N L 2 . Now the new value of number of turns required for the inductor is
  • 5. 5 8 104.0    FA Ll N c g new  (10) Now with new number of turns, the operating flux density can be obtained using equation (2) cnewf ac fANk V B 4 10  (11) Design a 250 V, 2.565 A single-phase AC inductor Applied voltage, V = 250 Volts Line current, I = 2.565 A Frequency = 50 Hertz. Current density, J = 300 Amp/cm2 Efficiency goal, 𝜂= 85 % Magnetic material = Silicon Magnetic material permeability, r = 1500 Flux density, Bac = 1.4 Tesla Window utilization, Ku = 0.4 Temperature rise goal, Tr = 50°C Step-1:- Calculation of apparent power VAVIP Lt 25.641565.2250  . Step-2:- Calculation of area product AP 4 44 94.171 3004.1504.044.4 1025.64110 cm JfBKk P A acuf t p       Step-3:- Selection of Core Core material suitable for designing the inductor is EI-150 for which the area product is 150.136 cm4 . Tables blow depicts the design and dimensional data for various laminations EI (Table 3.3 of Text Book).
  • 6. 6 Table 1: Design data for various EI laminations. Table 2: Dimensional data for various EI laminations. E D D Figure 3 Outline of EI laminations. For 𝑬𝑰 − 𝟏𝟓𝟎 lamination, Magnetic path length (MPL) = 22.9 cm Core weight = 2.334 Kg Copper weight = 853 gm
  • 7. 7 Mean length turn (MLT) = 22 cm Iron area, 𝐴 𝑐 = 13.79 cm2 Window area, 𝑊𝑎 = 10.89 cm2 Area product, 𝐴 𝑝 = 𝐴 𝑐 × 𝑊𝑎 = 150.136 cm4 Core geometry, 𝐾𝑔 = 37.579 cm5 Surface area, 𝐴 𝑡 = 479 cm2 Window length, G = 5.715 cm Lamination tongue, E = 3.81 cm Step-4:- Calculation of number of turns 584 79.13504.144.4 1025010 44       cacf fABk V N turns. Step-5: - Calculation of inductive reactance and inductance Inductive reactance,  466.97 565..2 250 L L I V X Inductance, H f X L L 31.0 502 466.97 2     Step-6:- Calculation of required air-gap mmcm MPL L AN l r c g 75.1175.0 1500 9.22 31.0 1079.135844.0104.0 8282                Step-7:- Calculation of Fringing flux factor 197.1 175.0 715.52 ln 79.13 175.0 1 2 ln1                  gc g l G A l F Step-8:- Calculation of new number of turns for the inductor 512 10197.179.134.0 175.031.0 104.0 88         FA Ll N c g new turns. Step-9:- Calculation of operating flux density with new turns 595.1 79.135051244.4 1025010 44       cnewf ac fANk V B T
  • 8. 8 Step-10:- Calculation of bare conductor area 22 )( 855.000855.0 300 565.2 mmcm J I A Bw  Step-11:- Selection of wire from wire table Cross-sectional area of the required wire close to the area calculated in step-10 is matching with the area of SWG-19 conductor. So the selected wire for designing the inductor is SWG-19. Table 3: Standard Wire Gauge table. Standard wire gauge Diameter in mm Cross-sectional area in mm2 Resistance per length in Ω/Km in µΩ/cm 18 1.22 1.17 14.8 148 19 1.02 0.811 21.3 213 20 0.914 0.657 26.3 263 For 19-SWG wire, bare conductor area if 0.811 mm2 and resistance is 213 cm/ . Step-12:- Calculation of inductor resistance Inductor resistance,   4.2102135122210213 66 newL NMLTR Step-13:- Inductor winding copper loss 79.154.2565.2 22  LL RIP Watts Step-14: - Calculation of Watts/Kg       K W 95.0595.150000557.0000557.0 86.168.186.168.1  acBf K W Step-15: - Calculation of core loss 22.2334.295.0  tfefe W K W P Watts Step-16: - Calculation of gap loss 146.13595.150175.081.3155.0 22  acgig fBElKP Watts Step-17: - Calculation of total losses 136.31146.132.279.15  gfecu PPPP Watts
  • 9. 9 Step-18: - Calculation inductor surface watt density 065.0 479 136.31   tA P  Watts/cm2 Step-19: - Calculation of temperature rise CTr 0826.0826.0 5.40065.0400400   Step-20:- Calculation of window utilization factor 38.0 887.10 00811.0512      a wBnew u w AN K