Eletronika

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Elektronika Kapal

Pendahuluan

V=V1+V2I=I1=I2I2I1V=V1+V2I=I1=I2I2I1Arus DC

V2V

V=V1=V2I=I1+I2V2I2I1V1IR2R1I

V

V=Voltage ; tegangan ; beda potensialTubuh kita hanya menerima 25 volt Klasifikasi tegangan:a) Extra High Voltage100kV-~b) High Voltage13,0kV-9,9kVc) Medium Voltage1,1kV-13,5kVd) Low Voltage0V-1kV

Di Jepang 110VIndonesia 220VSwedia lampu selalu menyala. Diesel=diesel generator dari nuklir (the most power full)

Electronica EngineeringSumber tegangan (battery)/generator, accuDistribusi

What is Electronics Eng?Electronics is distint from electrical and electromechanical scievice and technology.

It wich deals with the generation, distribution, switching, strorage, and corversion of electrical energy to and from other energy forms using wires, motors, generators, batteries, switches, relays, transformers, resistors, and other passive components.Rumus tahanan penghantar :

Switch, Relay, Transformator1 Phase (Transformator)

EMERGENCY SOURCES OF ELECTRICAL POWERSumber energinya adalah baterai (DC) dan Emergency genset (AC)Pertimbangan dalam perhitungan baterai :1. List kebutuhan daya pada saat emergency2. Rute pelayaran3. Perkiraan waktu pelayaranSumber energy listrik :DC : 1. Ampere Hour (Ah)2. Volt (V)AC :1. Va2. Watt3. V4. HzPeralatan-peralatan yang harus tetap bisa beroperasi pada saat emergency di kapal :1. Radar 2. Kompas, dllPertimbangan dalam perhitungan baterai :1. List kebutuhan daya pada saat emergency2. Rute pelayaran berkaitan dengan waktu pelayaran3. Perkiraan waktu pertolongan

Arus DC lebih berbahaya dari pada AC karena arusnya searah. Contoh gambar :

Solve for the current through R1 and R4 in the circuit of

Balanced CircuitMaka Gambar :Mencari RtotalRtotal = (R1 + R3) // (R2 + R4)(seri+ = (3 + 12) // (6 + 24)paralel) = 15 // 30= Rtotal = 10

Rkeseluruhan = Rs + Rtotal = 10 + 10 = 20 2.

Mencari I(arus)

R = Rtot = R5 + R = 10 +10 = 20Itot = V = I x R = 3 x 10 = 30 =

Series Parallel Circuit

E = 15V

V1 = 15V

Balanced // UnbalancedVab = () (O A) = OVI

But, Vab = Vad-Vbd = 0therefore, Vad-Vbd and

Which simplifies to

Bridge Circuit

Balanced Unbalnced

Open circuit close circuit

E = 60V

1. Req = (R1 // R2) + (R3//R4) = 10

RsReq

2. I.....?

V = IRRp1 = Rp2 = Rseri = 10Rtotal = 20I = IVR1 = IR1 x R1 = 2 x 3 = 6VR2 = IR2 x R2 = 1 x 6 = 6Series Paralel CircuitI = 120,78mA

8,6275V 24V 15,373V

R1 = 86,275mAR2 = 34,510mAR3 = 43,922mAR4 = 76,863mA

Bridge Circuit

2a. = . E = . E = = = = = 0.042

Gambar no 2.2. ()// () = (250)//(20000) = + = = = 246.91 ohm

Gambar no 1. = ()// ()= (50+200)//(5000)= (250)//(5000)= + = = 238.095 ohm= = = 0.042 A = = = 0.002 A = = 0.042 0.002= 0.040 A = = 0.04 x 50 = 2 v = = 0.002 x 5000= 10 v = -2 + 10= 8 v

Gambar 2. = 200 + 50= 250 ohmI = = = 0.004 A = I = 0.04 x 50= 2 v = 0 = - = -2 + 0= -2 v Gambar 3. = + = = 246.914 ohmI = = = 2 v = 2.5 v = - = -2 + 2.5= 0.5 v Soal!a. Gambarkan grafik output terhadap waktu untuk tegangan output nominal 220 volt AC 50 Hz b. Gambarkan grafik output terhadap waktu untuk tegangan output nominal 220 volt AC 60 Hz

Gambar

Total

E9999Volts

I0.94.5914.4Ampere

R10k2k1kxohms

= + I = = + + = = = R = = 14.4 mA = = = 4.5 = = = 9

Series -parallel circuitTentukan Requivalent = + + // = 5.182 ohm = + + // = 6.545 ohm = + + // = 8.727 ohm = + + // = 10.636 ohm

Example : = // 6 = 5.182 ohm1. Find current I2. Find

1. R. Equivalent2.

(1) = 0 (short circuit)(2)

(3)

= ( + ) // I = = = 0.042 A = 250 // 5 k ohm= + = = 238 ohm

Loop Analysis

Determinant for determinatorLoop I (15) I1 (6) I2 (3) I3 = 30 VLoop II -(6) I1 + (36) I2 (18) I3 = 0Loop III -(3) I1 (18) I2 + (24) I3 = 0

D = = 6264I1 = = = 2,586 AI2 = = = 0,948 AI3 = = = 1,034 AThe current trought R1 is found as IR1 = I1 I2 = 2,586 A 0,948 A = 1,638The current trought R2 is found as IR5 = I3 I2 = 1,034 A 0,948 A = 0,086 A to the right

Syntax matlab>>x=[15 -6 -3 ; -6 36 -18 ; -3 -18 24]

Convert to Y

R1 =

R2 =

R3 =

=

WAY DELTA CIRCUIT

60 30

90

R3=90 R2=60 R1=30 Berapa Req ?

R11 = 30 // 30 R13 = 30 // 90 R11 = 15 R13 = 22,5 R12 = 30 // 60 Req = R11 // (R12 + R13)R12 = 20 Req = 15 // (20 + 22,5)Req = 11,09

Rumus Untuk mengubah rangkaian Y ke

Untuk mengubah rangkaian ke Y

=

CONTOH SOAL

90 30 R3R21R1Tentukan nilai R1, R2, R3 !

60

Jawab!R1 = R3 = R1 = R3 = 10

R2 = R2 =

Tentukan Requivalen dari rangkaian dibawah ini.

30 10 10 15 10 10 30 V

Jawab!Ubah rangkaiannya menjadi lebih sederhana.

30 V25 40 20 o

Req = 20 + (25 // 40 )Req = 20 + 15,38 Req = 35, 38

Find the current through R6 for the current shown on figure:A. R6R1R3R5R4R2

10 V

R1 = 6 R4 = 3 R2 = 12 R5 = 18 R3 = 3 R6 = 6 Jawab!Pertama kita ubah dahulu bentuk Y

RBRARcR5R2R1

Cara mencari RAMencari RB

Mencari RC

Jadi apabila di gabung, rangkaiannya menjadi

R4R3RCRBRAR5

30 V

Hasil akhirnya menjadi:

RA + Rpar30 VReqR6

Mencari ReqReq = 2 + ((2+3) // (6+3))Req = 2 + 3,6 = 5,6 Maka, Rtot adalah: Rtot = Req + R6 Rtot= 5,6 + 6 = 11,6

Jadi, arus yang mengalir pada rangkaian tersebut adalah:Jika ingin mencari arus yang mengalir pada R3 dan R4

RAR6Rpar30 V

VRpar = Itot x RparVRpar = 2,586 x 3,6VRpar = 9, 324 V Tegangan pada rangkaian paralel

RC = 3 RB = 6 Vtot = 9, 324 V

R3 = 3 R4 = 3

Setelah itu kita cari IC3 dan IB4 Setelah itu, kita lanjutkan mencari V3, V4V3 = IC3 x R3V3 = 1,554 x 3 = 4,662 VV4 = IB4 x R4V4 = 1,023 x 3 = 3,046 V

Mencari V43, dimana...V43 = -V4 +V3V43 = -3,046 + 4,662 = 1,616 V

Jadi Arus yang mengalir pada R6 adalah...

Thevnin

Determine the Thvenin equivalent circuit external to the resistor RL for the circuit of Figure 910. Use the Thvenin equivalent circuit to calculate the current through RL.

1) Remove RL

2) Make terminal a & b

3) a

b

4) Vab I = 0

5) Vab E = 0

= 48 V

NETWORK THEOREMS

Example 9.2

R1 = 2,4 KE2 = 32 VR2 = 1,6 KE1 = 16 VR3 = 1,6 K I = 5 mA

kita buat rangkaiannya seperti dibawah ini :

= - 4.00 Volt

Kemudian rangkaian kita buat seperti demikian

RT = R1 // R2 // R3 = 0.6 k VR2(2) = ( 0.6 k ) (5 mA) = 300 V

Finally, R1 // R2 = 0,96 K

Example 9.5

R1 = 10 R4 = 50 R2 = 20 R3 = 20 E = 10 VoltR5 = 30

RTh = 10 // 20 + 20 // 50 = 6,67 + 14,29 = 20,95 Vab = =

AC FundamentalOutline1. Intoduction2. Generating AC voltage3. Voltage and Current convertion for AC4. Frequency, Period, Amplitude, and Peak Value5. Angular and graphic Relationship for sine wave6. Voltage and current asfnction of time7. Introdjuction to Pasor8. AC waveform and average value/effective value9. AC voltage and current measurementIntroduction

Triangle wave determine the current and its direction at t= 0,1,2,....12msIm= 4,5 . 10-3R = 20 . 10-3Vm = Im. R = 90 Volt

1 cycle per second

60 cycle per second = 60 Hz1 Hz = 1 cycle per second Frequency, Period, Amplitude, and Peak Value

Period T is the duration of one cycle measured in secondsExamplea) What is the period of a 50 HzAmplitude and Peak Value Amplitudo

Peak Value Basic sine wave equation

e = Em sin (volt)e = Em sin (volt)

1. Example if the amplitude of the waveform of figure 15-23(b) is Em =100 V, determine the coil voltage at 300 and 3300Answer : e1 = Em sin 1 = 100 sin 30 = 100. = 50 Volte2 = Em sin 2 = 100 sin 330 = 100. - = -50 Volt

Angular Velocity () = /t where 2 radians = 3600 radians = /1800 x degrees

2. Example:

= 2 f = .t2 = .T450 = x 450 = dengan rumus0,6 = x 0,6 = 1080

sin = V/Vm = sin-1 V/Vme = Em sin v = Vm sin i = Im sin Sinusoidal voltage and current is function of timeMaka

Voltage and current with phase ships

V = Vm sin (t + )V = Vm sin (t - )Phasor

Vm300 r RR 300Vm sin ( phasor) b. Resulting sin wave

Draw the phasor and waveform for current I = 25 sin t mA for t = 100 HzSolution : The phasor has a length of 25 mA and is drawn at its t = 0 position. Which is zero degrees as indicated in figure 15-36 since F = 100 Hz, t = 10 ms

25t (m/s)

= = 0,01 s = 10 ms

PhasorIf you plot v versus , you get the sine wave of figure 15-34 (b). Figure 15-35 illustrates the process. It shows snapshots of the magnitude Vm = 100 V rotating at = 300/s. for example, consider t = 0, 1, 2 and 3s.1. At 0 s, = 0, the phasor is at its 00 position, and its vertical projection is V=Vm sin t = 100 sin = 00. The point is at the origin.2. At t = 1 s, the phasor has rotated 300 and its vertical projection is V = 100 sin 300 = 50 V. This point is plotted at = 300 on the horizontal axis.3. At t = 2 s, Q = 600 and v = 100 sin 600 =87 V, which plotted at = 600 on the horizontal axis. Similarly at t = 3s, = 900, and v = 100 V. Continuing in this manner, the complete waveform is evolved.1. 0 sV = Vm sin t = 100 sin 0= 100 sin 0= 0 V2. 1 sV = Vm sin t = 100 sin 30. 1= 100 sin 30= 50 V1. 2 sV = Vm sin t = 100 sin 30. 2= 100 sin 60= 87 V

Figure of Fasor

t= 3st= 2s87V 100V100Vt= 1s50V

t= 0s100V 0V 300 600 900

Phase DifferenceIn phase v IResistive

Current leadsikapasitifv( i mendahului v)

Current lagsvinduktif ( v mendahului i ) i Difference phase

PHASE DIFFERENCE

I = Current Leands

Review Bilangan Kompleks

1. Z = a + bj a = bilangan real b = bilangan imanginer2. Z = R < 0 a = R . cos b = R . sin

soal : gambar dari e(t) = 200 sin (cot + 40) ?

jawab :

Thevenin dan NortonIf we wanted to find the current through the variable load resistor when RL = 0, RL= 2 and RL= 5K using exiting metode we would need to analyze.

1. Remove R2

2. Make terminal/node a and b

3. Set all source to zero

VAB>I=04.

VAB = VR2

5. VAB > E = 0

Vab = VR2lalu

30 VR2 = 2 R1 = 6

The thevenin resistance of the circuit is..VR2 = I x R2VR2 = 3,75 x 2VR2(b) = 7,5 V

VThevenin = VR2(a) + VR2(b)VThevenin = 3,75 +7,5 VThevenin = 11,25

RThevenin = R1 // R2RThevenin = 6 // 2RThevenin = 1,5 So IThevenin is..

Thevenin dan Norton (Lanjutan)

Langkah-langkah:1. Remove RL

RTH= 6 k // 2 k = 1,5 k2. Make terminal node

3. Set all source to zero

RAB= R1//R2 = 1,5 k

4. Vab => VR2

The thevenin resistance of the circuit is Vab E= 0

I= I = 3,75 x 10-3VR2 = I. R1Eth = VR2 x VR2= 3,75 x 2 = 7,5 Volt = 3,75 + 7,5 = 11,25Jadi, IL =

RESISTANCE AND SINUSOIDAL AC

I = IM sin t, V = VM sin t

IR = = = IM Sin t

VR1 = VM sin t,VR2 = VM Sin t,

Given : IR = 12 sin (t 180), R = 5 Asked : VR=?Solution : VR = IM .R {sin (t 180 )} = 12 . 5 {sin (t 180 )} = 60 sin (t 180)

INDUCTANCE AND SINUSOIDAL AC

VL = L . VL = L . IM sin t = L .IM cos tIf , VM = L . IMVL = VM cos tVL = VM sin (t + 900)

IV

090

90180

3535 + 90

120120 + 90

I lagsVI

900

0-90

18090

INDUCTIVE REACTANCEVM = L . IM = L () XL = L ()Maka, IM = VM = IM . XLIL = IM sin t

Example :Given: VL =100 sin (400t +700 ) and L= 0,2,asked: IL =? Solution: XL = . L = 400. 0,2= 80 , maka IM = = = 1,25 A

VI

7070-90= - 20

So, IL = IM sin (400t 20) IL = 1,25 sin(400t 20)

Circuit Analysis with Devices

The voltage across a 0,2 H inductance is Determine and sketch it.Given : Ask : ?Solution : rad/s therefore, The current lags the voltage by . Therefore, as indicated in figure.Inductance and Sinusoidal AC

Figure : with voltage at 70, current will be 90 later at -20.Example : ? rad/s Volt. Volt.

Inductance and Sinusoidal AC

Phase lag in an inductive circuit

Atau jadikan sin?

Apabila,I = 0V = +90

L = 0,2 HVL = 100 sin (400 +70)VIl?

Average Power

P = P = P = V . IVeff = Vm = Im = Latihan soal!V (t) = 100 sin tI (t) = 4 sin tP = (average power)P = = 200 voltInductive LoadPower to a purely inductive load. Energy stored during each quarter cycleis returned during the next quarter cycle. Average power is zero.V (t) = Vm sin(t + 900)I (t) = Im sin tP (t) =Vm Im cos t sin tP (t) = Vm Im ( sin 2 t)P (t) = sin 2 t

PL = V I sin 2 t(Reactive power Daya yang dikembalikan oleh loadnya)QL = V I (VAR)QL tidak mempunyai average power dan merupakan masalah yang sangat utama sistem operasi listrik.V = I XLQL = reactive powerI = V/XLQL = I2 XL = (VAR)

SOAL!

V (t) = Vm sin (t + 900)V (t) = 0I = I = = 5 AQL = I2 XLQL = 25 . 20QL = 500 VAR

Power to Capacitive Load

V (t) = Vm sin (t - 900)I (t) = Im sin tPC = Vm -cos t . Im sin tPC = -Vm Im cos t sin tPC = -Vm Im sin 2 t

Contoh Soal!1)

I = V / XLI = 100 / 40I = 2,5 AQC = I2 XCQC = (2,5)2 40QC = 250 VAR

2) For the RL circuit of figure 17.7 (a). I = 5 A. Determine P and Q

P = I2 RP = 52 3P = 75 wattQL = I2 XLQL = 52 4QL = 100 VAR

P = I2 RQC = I2 XC1P = (20)2 3QC = 400 . 6P = 1200 wattQC = 2400 VARQL = V2/XLQL = 4000 / 5QL = 800 VARQC = I2 XC2QC = 400 . 10QC = 4000 VAR

EVALUASI

I1EVALUASI I1. I5R4 = 200 ohmR3 = 350 ohmR2 = 250 ohmR1 = 100 ohmI4I2I3I6

24 V

Cari :A. VR1, VR2, VR3, VR4B. I1, I2, I3, I4, I5, I6Jawab!Rparalel 1Rparalel 2

Rtotal

Itotal = I1 = I4 (I1 =I4, karena pada rangkaian seri)

V1 = V2V3 = V4

Mencari nilai I1, I3, I5, dan I6

EVALUASI IIGanjil

R1 = 12 ohmGanjil1. ABR3 = 6 ohmR4 = 6 ohmR2 = 18 ohmR5 = 24 ohm15 V

Consider the bridge circuit of that figure:A. Is the bridge balance? Explain!B. Determine the voltage through R5C. Calculate the current across R5

Jawab!A. Jika seimbang, R1 x R4 = R2 x R3R1 x R4 = R2 x R312 x 6 = 18 x 672 ohm = 108 ohm, maka rangkaian tersebut tidak seimbang.

B dan C.Ubah rangkaian menjadi gambar dibawah ini

RA15 VR4R3

RBRC

Agar lebih mudah dan tidak membingungkan.

RA

Rparalel15 V

Mencari Nilai RTotalRTotal = RA + (( RC + R3 ) // ( RB + R4 ))RTotal = 4 + (( 5,333 + 6 ) // ( 8 + 6 ))RTotal = 4 + 6,26 = 10,26 ohmMencari nilai ITotalMencari Nilai VRparalelVRparalel = Rparalel + ITotalVRparalel = 6,26 + 1,46 = 9,152 VMencari nilai I3, I4, V3 dan V4

ABRBRCR3R4

I3 = VRparalel / ( R3 + RC )I4 = VRparalel / ( RB + R4 )I3 = 9,152 / ( 6 + 5,333 )I3 = 9,152 / ( 8 + 6 )I3 = 0,807 A I3 = 0,653 A

V3 = R3 x I3V4 = R4 x I4V3 = 6 x 0,807V4 = 6 x 0,653V3 = 4,842 VV4 = 3,918 V

B. VAB = -V4 + V3VAB = -3,918 + 4,842VAB = 0,924 V

C.

Genap1. R3 = 80 ohmR4 = 80 ohmR2 = 40 ohmR5=40ohmR1 = 30 ohm90 V

Consider the bridge circuit of that figure:A. Is the bridge balance? Explain!B. Determine the voltage through R5C. Calculate the current across R5

Jawab!A. Jika seimbang, R1 x R4 = R2 x R3R1 x R4 = R2 x R330 x 80 = 40 x 802400 ohm = 3200 ohm, maka rangkaian tersebut tidak seimbang.

15 VR4 = 80 ohmR3 = 80 ohmR5 = 40 ohmR2 = 40 ohmR1 = 30 ohm

Rangkaian sebelumnya kita ubah menjadi seperti dibawah ini

R2R3R1RaRb1Rc

Mencari nilai Ra, Rb, dan Rc

R3 = 80 ohmR4= 80 ohmRa = 10,909 ohm Rb=10,909 ohmRc = 14,545 ohm90 V

Mencari Nilai Hambatan pada rangkaian paralel diatas.Rparalel = ( Rb + R3 ) // ( Ra + R4 )Rparalel = ( 10,909 + 80 ) // ( 14,545 + 80 )Rparalel = 90,909 // 94,545Rparalel = 46,347 ohmMencari nilai hambatan total pada rangkaian tersebutRTotal = Ra + RParalelRTotal = 10,909 + 46,347RTotal = 57,256 ohmMencari arus yang mengalir pada tegangan tersebutMencari Nilai tegangan pada rangkaian paralel diatas.VRparalel = I x RparalelVRparalel = 1,572 x 46,347VRparalel = 72,857 VMencari arus dan tegangan yang mengalir pada R3 dan R4B. Determine the voltage through R5

C. Calculate the current across R5

EVALUASI III (Thevenin norton)Ganjil

R3 = 80 ohmR4 = 80 ohmR2 = 40 ohmR5=40ohmR1 = 30 ohm90 V1.

Ditanya: Carilah nilai I5 !Jawab!Langkah 1: Remove R5R3 = 80 ohmR4 = 80 ohmR2 = 40 ohmR1 = 30 ohm90 V

Langkah 2: Make terminal a dan bR3 = 80 ohmR4 = 80 ohmR2 = 40 ohmR1 = 30 ohm

BA

Langkah 3: Remove all sourceR3 = 80 ohmR4 = 80 ohmR2 = 40 ohmR1 = 30 ohm

R4 = 80 ohmR2 = 40 ohmR3 = 80 ohmR1 = 30 ohm

ABBA

Mencari Nilai RABRab = ( R1 // R3 ) + ( R2 // R4 )Rab = (21,818 + 26,667)Rab = 48,485 ohm

Langkah 4 : VAB

AB

Langkah 5

VbVaR4R2R3R1AB

Va = VR1 = Ia x R1 = 0,818 x 30 = 24,54 VVb = VR2 = Ib x R2 = 0,75 x 40 = 30 VVab = -Vb + VaVab = -30 + 24,53 = -5,46 VJadi nilai I5 atau Iab adalah...

Genap

R1 = 12 ohmGanjil2. R3 = 6 ohmR4 = 6 ohmR2 = 18 ohmR5 = 24 ohm15 V

Consider the bridge circuit of that figure:A. Is the bridge balance? Explain!B. Determine the voltage through R5C. Calculate the current across R5Jawab!Langkah 1: Remove R5

15 VR4 = 6 ohmR3 = 6 ohmR2 = 18 ohmR1 = 12 ohm

Langkah ke 2: Make Terminal a and b

15 VR4 = 6 ohmR3 = 6 ohmR2 = 18 ohmR1 = 12 ohm

BA

Langkah ke 3: Remove all sources

ABR4 = 6 ohmR2 = 18 ohmR3 = 6 ohmR1 = 12ohmBAR4 = 6 ohmR3 = 6 ohmR2 = 18 ohmR1 = 12 ohm

Mencari nilai R1 // R3 dan R2 // R4

Mencari nilai RABRAB = ( R1 // R3 ) + ( R2 // R4 )RAB = (4 + 4,5)Rab = 8,5 ohm

R4 = 6 ohmAR2 = 18 ohmBLangkah ke 4: VAB

15 VR3 = 6 ohmR1 = 12 ohm

Mencari IThevnin

Langkah 5

R4R2R3R1VbVa

Va = VR1 = Ia x R1 = 0,833 x 12 = 9,996 VVb = VR2 = Ib x R2 = 0,625 x 18 = 11,25 VV5 = Vab = -Vb + VaV5 = Vab = -11,25 + 9,996 = -1,254 VJadi nilai Iab atau I5 adalah...

EVALUASI IV

17.Given voltage . if , what is V at = 370?Jawab! 19. A sinusoidal voltage has a value of 50 V at = 1500 . whats Vm?Jawab! 20. Convert the following angles from radians to degress.a) Jawab! b) Jawab! c) d) Jawab! e) Jawab!

f) Jawab!

21.Conver the following angles from degree to radiansa) Jawab! b) Jawab! c) Jawab! d) Jawab! e) Jawab! f) Jawab!

22. A 50 KHz sine wave has an amplitude of 150 V. Sketch the wave for its axis scaled in microsecond.

Sketch the graphic

t-150155150V

23.if the period of the wave form on figure is 180 mS, complete current at t=30, 75, 140 and 315 mS

I50-50

Jawab!t = 30 mSt = 75 mSt = 140 mSt = 315 mS

25. a sinusoidal waveform has a period of 60 micro S and Vm = 80 V. Sketch waveform. What is its voltage at 4 micro S?Jawab!

Sketch the graphic

451560t (micro S)V80-80

ME091302 | ELEKTRONIKA KAPAL72