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EXPLAIN THE METHOD HOW TO CALCULATE WIRE'S DIAMETRE, WIRE'S LENGHT AND WINDING
Typology: Assignments
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Nama : Vania Kurnia Alvi (1310171021)
Kelas : 3 D4 Teknik Elektro Industri A
Tanggal : 21 April 2020
Dosen Pembimbing : Bapak Ir. Moh Zaenal Efendi, M.T.
The Buck Boost Converter has following parameters :
Vs(max) = 2 0 Volt
Vs(min) = 9 Volt
Vo = - 14 Volt
Io = 1 .5 A
R = 𝑉𝑜𝐼𝑜 = (^) 1.5^14 = 9.33 𝛺
Switching Frequency (fs) = 40 kHz
Components:
Q : MOSFET IRFP
D : MUR 1560 (Ultra Fast Recovery Diode)
Inductor (L) : Ferrit Core PQ 3535 with Cross sectional are (Ac=1.6 1 cm^2 );
Bobbin diameter (Dbob = 16.5 mm )
Rs : Snubber resistor ( 1 𝐾 Ohm ,5- 10 Watt)
Cs : Snubber Capacitor ( 5 nF , 1 KVolt )
Ds : Snubber diode ( FR3017)
The maximum inductor current
I𝐿(𝑚𝑎𝑥) = 𝐼𝐿(𝑎𝑣𝑔) +
Winding number of inductor
Bmax = 0.25 Tesla ; AC = in cm^2 = 1.6 1 cm^2 𝑛 = 𝐿×𝐼 𝐵𝑚𝑎𝑥𝐿(𝑚𝑎𝑥)×𝐴𝑐 104
= 0.18391 𝑚𝐻 × 10
−3×4. 0.25×1.61 10
4
= 19. 45 = 19
Wire Size is based on RMS current of inductor
2
2
Calculation of Wire Size
J = 4.5 A/mm^2 (current density)
Cross sectional Area of Wire (qw) 𝑞𝑤(𝑡) =
= 0.861 43214 mm^2
Diameter of Wire ( dw )
𝑑𝑤(𝑡) = √
= 1.04755 mm
Recalculate by assuming Number of Split Wire (∑ split = 9 ) o 𝐼𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 𝐼 ∑ 𝑠𝑝𝑙𝑖𝑡𝐿(𝑟𝑚𝑠)𝑡
o 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 =
𝐼𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 𝑗 =
= 0.0955 mm^2
o 𝑑𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √^4 𝜋 × 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡
= 0.348 mm ≈ 0.35 mm
Wire Size Diameter of bobbin PQ3535 (Dbob) = 16.5 mm = 1. 65 cm Circumference of Bobin (𝐾𝑏𝑜𝑏) = π × 𝐷𝑏𝑜𝑏 (𝐾𝑏𝑜𝑏) = π × 1. (𝐾𝑏𝑜𝑏) = 5.181 cm Total Wire Length = (𝑛(𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 × ∑ split) + 40% × (𝑛(𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 × ∑ split) Total Wire Length = (19 × 5.181 × 9) + 40% × (19 × 5.181 × 9)
Lampiran
1. Data sheet IRFP
