Tugas Workshop Konverter Daya Spwm Inverter Unipolar: Dosen: Ir. Moh. Zaenal Efendi, Mt

TUGAS WORKSHOP KONVERTER DAYA SPWM INVERTER UNIPOLAR

DOSEN: Ir. Moh. Zaenal Efendi, MT

Disusun Oleh: Ahmad Daerobi

(1303181035)

Elsa Firul A

(1303181036)

Moh. Ilham Fadillah

(1303181041)

Mochammad Irfan A

(1303181048)

2 D3 Elektro Industri B

PROGAM STUDI D3 TEKNIK ELEKTRO INDUSTRI DEPARTEMEN TEKNIK ELEKTRO POLITEKNIK ELEKTRONIKA NEGERI SURABAYA 2020

Tugas 3 Soal :

Jawaban : 1. Diketahui:  f = 50Hz  Vdc = 100v  ma =1  mf = 16  R = 100Ω  L = 100mH Ditanya : A. Nilai tegangan dan arus pada frekuensi fundamental B. Vo (rms) C. Io (rms) D. THDv E. THDi F. Daya aktif yang diserap beban resistor G. Daya semu yang diserap impedansi Penyelesaian : A. Nilai Amplitude  Tegangan V1 = ma. Vdc = 1 .100 = 100 V V1 (rms) = 70,71 V



Arus

I1

=

100 √1002 +(1.2𝜋.50.0,1)2

= 0,954 A I1 (rms) = 0,674 A Dengan mf = 16, Harmonic pertama pada n = 31, 33,39 and 45 V(31) = V(33) = (0,18) x (100)

= 18 V

V(31) (rms)

= 12,727 V

= V(33) (rms)

V(29) = V(35) = (0,21) x (100)

= 21 V

V(29) (rms)

= 14,849 V

I(31) =

= V(35) (rms) 18

√1002 +(31.2𝜋.50.0,1)2

= 0,018 A

Irms = 0,01272 A I(33) =

18 √1002 +(33.2𝜋.50.0,1)2

= 0,017 A

Irms = 0,01202 A I(29)=

21 √1002 +(29.2𝜋.50.0,1)2

= 0,022 A

Irms = 0,01555 A I(35) =

21 √1002 +(35.2𝜋.50.0,1)2

=0,019 A

Irms = 0,013435 A B. Vo (rms) 𝑉𝑜 (𝑟𝑚𝑠) = √𝑉1(𝑟𝑚𝑠)2 + 𝑉31(𝑟𝑚𝑠)2 + 𝑉33(𝑟𝑚𝑠)2 + 𝑉29(𝑟𝑚𝑠)2 + 𝑉35(𝑟𝑚𝑠)2 𝑉𝑜 (𝑟𝑚𝑠) = √(70,71 )2 + (12,727 )2 + (12,727 )2 + (14,849 )2 + (14,849 )2 𝑉𝑜 (𝑟𝑚𝑠) = 75,926 𝑉 C. Io (rms) 𝐼𝑜 (𝑟𝑚𝑠) = √𝐼1(𝑟𝑚𝑠)2 + 𝐼31(𝑟𝑚𝑠)2 + 𝐼33(𝑟𝑚𝑠)2 + 𝐼29(𝑟𝑚𝑠)2 + 𝐼35(𝑟𝑚𝑠)2 𝐼𝑜 (𝑟𝑚𝑠) = √(0,674 )2 + (0,01272)2 + (0,01202)2 + (0,01555)2 + (0,013435)2 𝐼𝑜 (𝑟𝑚𝑠) = 0,6745 𝐴

D. THDv THDv = =

√ 𝑉31(𝑟𝑚𝑠)2 + 𝑉33(𝑟𝑚𝑠)2 + 𝑉29(𝑟𝑚𝑠)2 + 𝑉35(𝑟𝑚𝑠)2 𝑉1 𝑟𝑚𝑠 √(12,727 )2 + (12,727 )2+ (14,849 )2 + (14,849 )2 70,71

= 0,391 E. THDi THDi =

=

√ 𝐼31(𝑟𝑚𝑠)2 + 𝐼33(𝑟𝑚𝑠)2 + 𝐼29(𝑟𝑚𝑠)2 + 𝐼35(𝑟𝑚𝑠)2 𝐼1 𝑟𝑚𝑠 √(0,01272)2 + (0,01202)2+ (0,01555)2+ (0,013435)2 0,674

= 0,04 F. Daya aktif yang diserap beban resistor Po = 𝐼1(𝑟𝑚𝑠)2 𝑥 𝑅 + 𝐼31(𝑟𝑚𝑠)2 𝑥 𝑅 + 𝐼33(𝑟𝑚𝑠)2 𝑥 𝑅 + 𝐼29(𝑟𝑚𝑠)2 𝑥 𝑅 + 𝐼35(𝑟𝑚𝑠)2 𝑥 𝑅 = (0,674)2 𝑥 100 + (0,01272)2 𝑥 100 + (0,01202)2 𝑥 100 + (0,01555)2 𝑥100 + (0,013435)2 𝑥100 = 45,5 W G. Daya semu yang diserap impedansi Z(n)

= √𝑅2 + (2𝜋𝑓(𝑛)𝐿)2

Z(1)

= √𝑅2 + (2𝜋𝑓(1)𝐿)2 = √1002 + (2𝜋𝑥50𝑥0,1)2 = 104,818 Ω

Z(31)

= √𝑅2 + (2𝜋𝑓(31)𝐿)2 = √1002 + (31𝑥2𝜋𝑥50𝑥0,1)2 = 979,01 Ω

Z(33)

= √𝑅2 + (2𝜋𝑓(33)𝐿)2 = √1002 + (33𝑥2𝜋𝑥50𝑥0,1)2 = 1041,53 Ω

Z(29)

= √𝑅2 + (2𝜋𝑓(29)𝐿)2 = √1002 + (29𝑥2𝜋𝑥50𝑥0,1)2 = 916,533 Ω

Z(35)

= √𝑅2 + (2𝜋𝑓(35)𝐿)2 = √1002 + (35𝑥2𝜋𝑥50𝑥0,1)2 = 1104,095 Ω

So = 𝐼(1)𝑟𝑚𝑠 2 𝑥 𝑍1 + 𝐼(31)𝑟𝑚𝑠 2 𝑥 𝑍31 + 𝐼(33)𝑟𝑚𝑠 2 𝑥 𝑍33 + 𝐼(35)𝑟𝑚𝑠 2 𝑥 𝑍35 + 𝐼(29)𝑟𝑚𝑠 2 𝑥 𝑍 29 2

2

2

So = (0,674) 𝑥 104,818 + (0,01272) 𝑥 979,01 + (0,01202) 𝑥 1041,53 2

2

+ (0,013435) 𝑥 1104,095 + (0,01555) 𝑥 916,533 So = 48,334 𝑉𝐴

Table Fourier Series Quantities for the PWM Inverter n

f (HZ)

Zn(Ω)

In(A)

In,rms(A)

Pn(W)

Sn(VA)

1

50

104,818

0,954

0,674

45,427

47,616

29

1450

916,533

0,022

0,01555

0,024

0,221

31

1550

979,01

0,018

0,01272

0,016

0,158

33

1650

1041,53

0,017

0,01202

0,0144

0,15

35

1750

1104,095

0,019

0,013435

0,018

0,199

2. Simulasi

𝑉𝑟𝑒𝑓 𝑉𝑐𝑎𝑟𝑟 𝑉𝑟𝑒𝑓 = 𝑉𝑐𝑎𝑟𝑟

ma = 1

Vref = 4 𝑉𝑃𝑃 VCarr = 4 𝑉𝑃𝑃 𝑓𝑐𝑎𝑟𝑟 𝑓𝑟𝑒𝑓 𝑓𝑐𝑎𝑟𝑟 16 = 50 fcarr = 800Hz mf =

V Carrier dan V Sinus

S1

S3

V SPWM

V (Out)

I (Out)

Active Power

Apparent Power

THDv

1. V1

2. V29

3. V31

4. V33

5. V35

THDi

1. I 1

2. I 29

3. I 31

4. I 33

5. I 35