Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

HOW TO CALCULATE WIRE'S DIAMETRE, WIRE'S LENGHT AND WINDING, Assignments of Power Electronics

EXPLAIN THE METHOD HOW TO CALCULATE WIRE'S DIAMETRE, WIRE'S LENGHT AND WINDING

Typology: Assignments

2019/2020

Uploaded on 05/06/2020

vania-kurnia-alvi
vania-kurnia-alvi 🇮🇩

4.8

(4)

5 documents

1 / 12

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
PROJECT WORK BUCK BOOST CONVERTER 2020
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.
POLITEKNIK ELEKTRONIKA NEGERI SURABAYA
SURABAYA
2020
pf3
pf4
pf5
pf8
pf9
pfa

Partial preview of the text

Download HOW TO CALCULATE WIRE'S DIAMETRE, WIRE'S LENGHT AND WINDING and more Assignments Power Electronics in PDF only on Docsity!

PROJECT WORK BUCK BOOST CONVERTER 2020

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.

POLITEKNIK ELEKTRONIKA NEGERI SURABAYA

SURABAYA

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)

𝐿 = 1. 8391 × 10 −^4 H

The maximum inductor current

I𝐿(𝑚𝑎𝑥) = 𝐼𝐿(𝑎𝑣𝑔) +

I𝐿(𝑚𝑎𝑥) = 3.87 +

I𝐿(𝑚𝑎𝑥) = 4.257 A

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

𝐼𝐿(𝑟𝑚𝑠)𝑡 = √(𝐼𝐿(𝑎𝑣𝑔))^2 + (

2

𝐼𝐿(𝑟𝑚𝑠)𝑡 = √(3.87)^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) = √

𝜋 × 𝑞𝑤(𝑡)

𝑑𝑤(𝑡) = √^

3.14 × 0.

= 1.04755 mm

 Recalculate by assuming Number of Split Wire (∑ split = 9 ) o 𝐼𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 𝐼 ∑ 𝑠𝑝𝑙𝑖𝑡𝐿(𝑟𝑚𝑠)𝑡

𝐼𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 0.43071607 A

o 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 =

𝐼𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 𝑗 =

= 0.0955 mm^2

o 𝑑𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √^4 𝜋 × 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡

= √^

3.14 ×^0.^0955

= 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)

 𝑅𝑆 < 2×𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟𝐷𝑇

0.61 × 40 × 10^13

2 × 4.822 × 10−

Lampiran

1. Data sheet IRFP

    1. Data Sheet MUR