Workshop Konverter Daya Flyback Converter Assigment 2: Dosen: Ir. Moh. Zaenal Efendi, Mt

WORKSHOP KONVERTER DAYA FLYBACK CONVERTER ASSIGMENT 2

Dosen : Ir. Moh. Zaenal Efendi, MT.

Disusun oleh : Iriana Kusuma Dewi (1303181003) Muhammad Reza Naufal H. (1303181010) Dane (1303181016)

TEKNIK ELEKTRO INDUSTRI POLITEKNIK ELEKTRONIKA NEGERI SURABAYA TAHUN 2019/2020

SOAL : 1. The flybackconverter of Fig. 7-2 has an input of 44 V, an output of 3 V, a duty ratio of 0.32, and a switching frequency of 100 kHz. The load resistor is 1 . (a) Determine the transformer turns ratio. (b) Determine the transformermagnetizing inductance Lm such that the minimum inductor current is 20 percent of the average. (c) capacitor to limit the ripple to less than 0.25 percent. 2. Design a flybackconverter for an input of 24 V and an output of 40 W at 40 V. A switching frequency of 40 kHz. Specify the transformer turns ratio and magnetizing inductance for CCM operation, and capacitor to limit the ripple to less than 0.5 percent. JAWABAN : 1. Diketahui : Vs = 44V Vo = 3V f = 100kHz R = 1ohm D = 0,32 r = 0.25% Jawab : a. Determine the transformer turns ratio.  D  N 2  Vo  Vs     1  D  N1   0,32   N 2  3  44     1  0,32   N1 

 N2  0,14     N1  b. Determine the transformermagnetizing inductance Lm such that the minimum inductor current is 20 percent of the average. Vo  N 2  ILm =   (1  D) R  N1  =

3  0,14  (1  0,32)1

= 0,63 A IL = 20%xIL

= 20% x 0,61 = 0,12A

Lm

=

1 1 xVsxDx f IL

=

1 1 x44 x0,32 x 100.000 0,12

= 1,17mH c. Capacitor to limit the ripple to less than 0.25 percent. D C = Rxfxr 0,32 = 1x100.000 x0,0025 = 1,28 mF 2. Diketahui; Vs = 24V Vo = 40V Po = 40W F = 40kHz r = 0.5% Jawab : Untuk menghitung L dan C digunakan duty cycle 50% Mencari nilai R : Vo 2 Po  R Vo 2 R Po 402 R 40 R  40 Mencari rasio transformator : 𝐷 𝑁2 𝑉𝑜 = 𝑉𝑠 ( ) 1 − 𝐷 𝑁1 0,5 𝑁2 40 = 24 ( ) 1 − 0,5 𝑁1 𝑁2 5 = 𝑁1 3

Mencari nilai ILm : Vo  N 2  ILm =   (1  D) R  N1  =

40 5   (1  0,5)40  3 

= 3,34 A Mencari nilai ∆𝐼𝐿 : IL

= 20%xIL = 20% x 3,34 = 0,668A

Mencari nilai Lm : Lm

=

1 1 xVsxDx f IL

=

1 1 x24 x0,5 x 40.000 0,668

= 0,449mH Mencari nilai C : C

=

D Rxfxr

0,5 40 x40.000 x0,005 = 0,0625 mF = 62,5uF =