The document describes a boost converter project including specifications and component selection. It specifies an input voltage range of 24-30V, output of 60V and 2.5A. Components selected include a MOSFET, diode, ferrite-core inductor with 41 turns, 625uF output capacitor, and 1kΩ 1W snubber resistor with 3.02nF capacitor. Inductor calculations determine a value of 291.764uH is needed with 56m of 0.35mm wire required.

1. PROJECT WORK BOOST CONVERTER 2018 – 2019
Group Member (3 – D4 ELIN A) :
1. Abdillah Aziz M 1310161018
2. Mohammad Agung D 1310161024
3. Gilang Andaru T 1310161027
The boost converter has following parameters :
𝑉𝑠(𝑚𝑎𝑥) = 30 Volt
𝑉𝑠(𝑚𝑖𝑛) = 24 Volt
𝑉𝑜 = 60 Volt
𝐼𝑜 = 2.5 Ampere
Switching Frequency (𝑓𝑠) = 40 kHz
Components :
Q : MOSFET IRF460
D : MUR 1560 (Ultra Fast Recovery Diode)
Inductor (L) : Ferrit core PQ 3535 with Cross sectional are (𝐴 𝑐 = 1.96 𝑐𝑚2
);
Bobbin diameter ( 𝐷 𝑏𝑜𝑏 = 17 𝑚𝑚)
𝑅 𝑠 : Snubber resistor (..... Ohm, 5 – 10 watt)
𝐶𝑠 : Snubber resistor (..... nF, 1 kVolt)
𝐷 𝑠 : Snubber diode (FR3017)

2. SOLUTION :
 Duty Cycle :
𝐷 = 1 − (
𝑉 𝑠(𝑚𝑖𝑛 )
𝑉𝑜
)
𝐷 = 1 − (
24
60
)
𝐷 = 1 − (
2
5
) 0.6 = 60%
 The inductor Value :
𝑅 =
𝑉𝑜
𝐼𝑜
=
60
2.5
= 24 Ω
𝐼𝐿( 𝑎𝑣𝑔) =
𝑉𝑠
(1 − 𝐷) 2 𝑅
=
24
(1 − 0.6) 2 × 24
= 6.25 𝐴𝑚𝑝𝑒𝑟𝑒
∆𝐼𝐿 = 20% × 𝐼𝐿( 𝑎𝑣𝑔) = 0.2 × 𝐼𝐿( 𝑎𝑣𝑔) = 0.2 × 6.25 = 1.25 𝐴𝑚𝑝𝑒𝑟𝑒
𝐿 = (
1
𝑓
) × [( 𝑉𝑜 + 𝑉𝐹 ) − 𝑉𝑠(𝑚𝑖𝑛)]× (
𝑉𝑠(𝑚𝑖𝑛)
( 𝑉𝑜 + 𝑉𝐹 )
) × (
1
∆𝐼𝐿
)
𝐿 = (
1
40 × 103
) × [(60 + 1.2) − 24] × (
24
(60 + 1.2)
) × (
1
1.25
)
𝐿 = 2.5 × 10−5
× 37.2 × 0.3921568 × 0.8 = 2.91764× 10−4
= 291.764 𝜇𝐻
 The maximum inductor Value :
𝑖 𝐿( 𝑚𝑎𝑥) = 𝐼𝐿( 𝑎𝑣𝑔) +
∆𝐼𝐿
2
= 6.25 +
1.25
2
= 6.25 + 0.625 = 6.875 𝐴𝑚𝑝𝑒𝑟𝑒
 Winding number of inductor :
𝑛 =
𝐿 × 𝑖 𝐿( 𝑚𝑎𝑥)
𝐵 𝑚𝑎𝑥 × 𝐴 𝑐
× 104
=
2.917647058 × 10−4
× 6.875
0.25 × 1.96
× 104
= 40.93637 ≈ 41
 Winding number of inductor :
𝑖 𝐿( 𝑟𝑚𝑠) 𝑡 = √(𝑖 𝐿( 𝑎𝑣𝑔))2 + (
∆𝐼 𝐿
2
√3
)
2
= √6.252 + (
1.25
2
√3
)
2
= 6.260408001Ampere
 Calculation of Wire Size :
 Cross Sectional Area of Wire (𝑞 𝑤)
𝑞 𝑤(𝑡) =
𝑖 𝐿( 𝑟𝑚𝑠) 𝑡
𝐽
=
6.260408001
4.5
= 1.391201778 𝑚𝑚2
 Diameter of Wire (𝑑 𝑤)
𝑑 𝑤(𝑡) = √
4
𝜋
× 𝑞 𝑤(𝑡) = √
4
𝜋
× 1.391201778 = 1.330914392 𝑚𝑚

3.  Recalculate by assuming number of split wire (∑Split) = 15
𝑖 𝐿( 𝑟𝑚𝑠) 𝑠𝑝𝑙𝑖𝑡 =
𝑖 𝐿( 𝑟𝑚𝑠) 𝑡
∑Split
=
6.260408001
15
= 0.41736 𝐴𝑚𝑝𝑒𝑟𝑒
𝑞 𝑤( 𝑡) 𝑠𝑝𝑙𝑖𝑡 =
𝑖 𝐿( 𝑟𝑚𝑠) 𝑠𝑝𝑙𝑖𝑡
𝐽
=
0.41736
4.5
= 0.0927467852 𝑚𝑚2
≈ 0.1𝑚𝑚2
𝑑 𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √
4
𝜋
× 𝑞 𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √
4
𝜋
× 0.1 = 0.35 𝑚𝑚
 Wire Size (∑𝑺𝒑𝒍𝒊𝒕 𝑪𝒉𝒐𝒐𝒔𝒆 ) = 15+3
Diameter of bobbin PQ3535 ( 𝑑 𝑏𝑜𝑏) = 17 𝑚𝑚 = 1.7 𝑐𝑚
Circumference of Bobin ( 𝐾𝑏𝑜𝑏) = 𝜋 × 𝑑 𝑏𝑜𝑏 = 𝜋 × 1.7 = 5.340707511 cm
Total Wire Length = (𝑛( 𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 × ∑Split) + 40% × (𝑛( 𝑤𝑖𝑛𝑑𝑖𝑛𝑔) 𝜋 × 𝐾𝑏𝑜𝑏 ×
∑Split)
Total Wire Length = (41 × 5.341 × 18) + [40% × (41 × 5.341 × 18)]
Total Wire Length = (3941.65)+ [1576.66] = 5518.31 𝑐𝑚 = 55.18𝑚 ≈ 56 𝑚
 Output Capacitance
𝐶 𝑜 =
𝑉𝑜 × 𝐷
𝑅 × ∆𝑉𝑜 × 𝑓
=
60 × 0.6
24 × 0.001 × 60 × 40 × 103
= 6.25 × 10−4
𝐹 = 625 𝜇𝐹
≈ 1000 𝜇𝐹,100 𝑉𝑜𝑙𝑡
 Snubber Circuit
𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 ≈
𝐼 𝑂𝑁 × 𝑡𝑓𝑎𝑙𝑙
2 × 𝑉𝑂𝐹𝐹
𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 =
𝐼 𝑂 ×
𝑉𝑜
𝑉𝑆
× 𝑡𝑓𝑎𝑙𝑙
2 × 𝑉𝑂
=
2.5 ×
60
24
× 58 × 10−9
2 × 60
= 3.020833333× 10−9
𝐹
= 3.02 𝑛𝐹
𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 𝐶ℎ𝑜𝑜𝑠𝑒 ≈ 4.7𝑛𝐹, 1 Kvolt
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 <
𝐷𝑇
2 × 𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 <
0.6 × 2.5 × 10−5
2 × 3.020833333 × 10−9
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 < 2482.758621 Ω
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 𝐶ℎ𝑜𝑜𝑠𝑒 =
1
2
× 2482.758621 = 1241.379311 Ω ≈ 1 𝐾Ω, 10 watt