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DESIGN OF FLYBACK CONVERTER, Assignments of Power Electronics

Institut Teknologi Sepuluh Nopember (ITSN)Power Electronics

Includes Parameter Calculation in order to designing Flyback Converter

Typology: Assignments

2019/2020

Uploaded on 05/14/2020

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DESIGN OF FLYBACK CONVERTER (A)
PARAMETER CALCULATION
Name : Vania Kurnia Alvi
NRP : 1310171021
Class : 3 D4 Teknik Elektro Industri A
Date : 15 Mei 2020
Lecture : Bapak Ir. Moh Zaenal Efendi, M.T.
POLITEKNIK ELEKTRONIKA NEGERI SURABAYA
SURABAYA
2020
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DESIGN OF FLYBACK CONVERTER (A) PARAMETER CALCULATION

Name : Vania Kurnia Alvi

NRP : 1310171021

Class : 3 D4 Teknik Elektro Industri A

Date : 15 Mei 2020

Lecture : Bapak Ir. Moh Zaenal Efendi, M.T.

POLITEKNIK ELEKTRONIKA NEGERI SURABAYA

SURABAYA 2020

PROJECT WORK – 2020

DESIGN OF FLYBACK CONVERTER (A)

Design a flyback converter from these parameters and solutions

  • Vs(min) = 80 Volt
  • Vs(max) = 90 Volt
  • Vo = 20 Volt
  • Io = 2,5 A
  • Swicthing frequency ( fs ) = 40 kHz
  • Duty cycle = Dmax = 0.

Component: Q : MOSFET IRFPG D : Fast Recovery Diode (STTH60L06C) L : Ferrit Core PQ 3535 with Cross sectional are (Ac=1.96 cm2) Co : Output capacitor (Co=? ;50Volt-100Volt) Rs : Snubber resistor (Rs=? 10 (20) watt) Cs : Snubber capacitor (Cs=? 1kVolt) Ds : Snubber diode (FR307)

  1. DC Current (Idc), ΔI, Vro

 𝐼𝑑𝑐 = (^) 𝑉𝑆(𝑚𝑖𝑛𝑃)𝑖𝑛. 𝐷𝑚𝑎𝑘𝑠

 𝐼𝑑𝑐 = (^) ( 8055 ×,^56 0. 5 )

 𝐼𝑑𝑐 = 5540 ,^56

 𝐼𝑑𝑐 = 1 , 389 𝐴

 𝐼𝑑𝑐 ≈ 1 , 39 𝐴

 ∆I =

( 𝑉𝑆(𝑚𝑖𝑛) (^). 𝐷𝑚𝑎𝑘𝑠 ) 𝐿𝑚. 𝑓𝑠

 ∆I = (^719) , 94 (^.^8010 ×− (^6 0) ×,^5 40 ). 103

 ∆I = (^28) ,^407976

 ∆I = 1. 389

 ∆I ≈ 1 , 39 𝐴

 𝑉𝑟𝑜 = 1− 𝐷𝐷𝑚𝑎𝑘𝑠𝑚𝑎𝑘𝑠 × 𝑉𝑆(𝑚𝑖𝑛)

 𝑉𝑟𝑜 = 1− 0,50,5 × 80

  1. The Maximum Inductor Current (Imax)

 𝐼𝑚𝑎𝑥 = 1,12 × ( 𝐼𝑑𝑐 + ∆ 2 I )

 𝐼𝑚𝑎𝑥 = 1,12 × ( 1,39 + 1.39 2 )

 𝐼𝑚𝑎𝑥 = 1,12 × 2,

 𝐼𝑚𝑎𝑥 = 2,

 𝐼𝑚𝑎𝑥 ≈ 2,34 𝐴

  1. Ratio of transformator (n)

 𝑛 = N Nps = (^) VoV+ VroF

 𝑛 = (^) 20 +1,3^80

 𝑛 = 3 , 755868

 𝑛 ≈ 3 , 76

  1. Winding number of inductor

𝐵𝑠𝑎𝑡 =

𝐴𝑐 × n

𝐵𝑠𝑎𝑡 =

1,96 × 3,

 𝑁𝑝(𝑚𝑖𝑛) = L Bm .satImax.Ac 104 ; Bmax = 0,25 T; Ac = 1 , 96 mm^2

 𝑁𝑝(𝑚𝑖𝑛) = 719,94.

−6 (^) × 2, 0,135692574 ×1,96 10

4

 𝑁𝑝(𝑚𝑖𝑛) = (^) 0,26595744516,

 𝑁𝑝(𝑚𝑖𝑛) = 63 , 34320138

 𝑁𝑝(𝑚𝑖𝑛) ≈ 63 , 34

  1. Primary Winding

 𝑁𝑝 = 2. 𝑁𝑝(𝑚𝑖𝑛)

 𝑁𝑝 = 2 × 63 , 34

 𝑁𝑝 = 126 , 68

 𝐶 = 𝑅×𝐷 𝑓××^ 𝑉 ∆𝑜𝑉𝑜

 𝐶 = 8 × 40. 100 , 35 ×× 020 , 001 × 20

 𝐶 = 1. 562 , 5 μF

  1. Size of wire (TARGET Size of wire = 0.4 mm)

Split Wire (∑ split = 2) ; J = 4.5 A/mm^2 (current density)

o 𝐼𝑝(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 𝐼𝑝(𝑟𝑚𝑠)𝑡 ∑ 𝑠𝑝𝑙𝑖𝑡

𝐼𝑝(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 0. 725 A

o 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = 𝐼𝑝(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 𝑗

= 0.16 mm^2

o 𝑑𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √𝜋^4 × 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡

= √^

3.14 ×^0.^16

= 0.45 mm

Split Wire (∑ split = 10 ) ; J = 4.5 A/mm^2 (current density) o 𝐼𝑠(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 𝐼𝑠(𝑟𝑚𝑠)𝑡 ∑ 𝑠𝑝𝑙𝑖𝑡

𝐼𝑠(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 0. 545 A

o 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = 𝐼𝑠(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 𝑗

= 0.12 mm^2

o 𝑑𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √𝜋^4 × 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡

= √^

× 0. 12

= 0.39 mm ≈ 0.4 mm

  1. Length of wire
    • diameter of bobbin : Dbob = 17 mm = 1,7 cm Pp = (Np × Kbobin × Σsplit) + 50% Ps = (Ns × Kbobin × Σsplit) + 50% Circumference of Bobin (𝐾𝑏𝑜𝑏) = π × 𝐷𝑏𝑜𝑏

(𝐾𝑏𝑜𝑏) = π × 1, (𝐾𝑏𝑜𝑏) = 5,338 cm

c. Ion

 𝐼𝑜𝑛 = 𝐼𝐿𝑀 = 𝑉𝑜

2 𝑉(max) × 𝐷 × 𝑅

 𝐼𝑜𝑛 = 20

2 90 × 0 , 5 × 8

 𝐼𝑜𝑛 = (^400360)

 𝐼𝑜𝑛 = 1 , 11 𝐴

d. Voff

 𝑉𝑜𝑓𝑓 = 𝑉𝑠 + 𝑉𝑜 (𝑁 𝑁^1 2

 𝑉𝑜𝑓𝑓 = 90 + 20 (^82 )

e. Csnubber

 𝐶𝑠 ≈ 𝐼𝑂𝑁 2×𝑉×𝑡𝑂𝐹𝐹𝑓𝑎𝑙𝑙

 𝐶𝑠 = 1,11×30. − 2×  𝐶𝑠 = 0,097941176 × 10−9^ 𝐹

 𝐶𝑠 = 0,09 𝑛𝐹 𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 𝑐ℎ𝑜𝑜𝑠𝑒 ≈ 0,1 𝑛𝐹, 1𝐾𝑉olt

f. Rsnubber

 𝑅𝑆 < (^) 2×𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟𝐷𝑇

 𝑅𝑆 < 0.5 ×^

1 40×10^3 2×0,1×10−  𝑅𝑆 < 62.500 Ω  𝑅𝑆 < 62,5 𝐾Ω  𝑅𝑆 ≈ 30 𝐾Ω  𝑅𝑠𝑛𝑢𝑏𝑏𝑒𝑟 𝑐ℎ𝑜𝑜𝑠𝑒 ≈ 30 𝐾Ω, 10 𝑊𝑎𝑡𝑡