Dasar Dasar Teknik Teknik Elekt Elektro ro
EL.112 (3 sks) Enjang A.Juanda/ Lukmanul Hakim
Silabus Mata Kuliah Secara garis besar disajikan: 1. Peng Pengantar antar T Teknik eknik Elek Elektro. tro. 2. Dasar-das Dasar-dasar ar rrangka angkaian ian listri listrik. k. 3. Resp Respon onkondisi ran rangkaia gkaian n bola bolak-bali k-balikk pad padaa steady state. 4. Pe Penga nganta ntarr sys system tem.. 5. Das Dasar ar elektr elektroni onika. ka. 6. Dasa Dasarr komp komponen onen elek elektroni tronika ka semikonduktor. 7. Peng Pengantar antar anali analisa sa jari jaringan. ngan. 8. Dasar Dasar el elektro ektronika nika digital digital & mikroprosesor. 9. Pe Peng ngua uatt OPOP-Am Amp. p.
Tujuan Setelah selesai mengikuti mata kuliah ini mahasiswa diharapkan mampu menjelaskan dasar teknik elektro d a n sedapat mungkin mempraktekkan bagianbagian yang praktisnya tentang dasar t e k n i k e l e k t r o . Evaluasi - Kehadiran - Tugas Presentasi dan diskusi - Makalah - UTS - UAS
Rincian Bahan I). Membahas silabus perkuliahan dan mengakomodasikan berbagai masukan dari mahasiswa untuk memberi kemungkinan revisi terhadap pokok bahasan yang dianggap tidak penting dan memasukkan pokok bahasan yang dianggap penting. Sesuai dengan apa yang dikemukakan dalam silabus, pada pertemuan ini dikemukakan pula tujuan, ruang lingkup, prosedur perkuliahan, penjelasan tentang tugas yang harus dilakukan mahasiswa, ujian yang harus diikuti termasuk jenis soal dan cara menyelesaikan/ menjawab pertanyaan, dan sumber-sumber. sumber-sumber. Terakhir, menyampaikan uraian pendahuluan tentang DasarTeknik Elektro/ Pengantar Teknik Elektro. II). Pengertian dan definisi-definisi yang terkait dengan dasar teknik elektro. III). Rangkaian-rangkaian listrik dasar: DC dan sistem DC. IV). Rangkaian-rangkaian listrik dasar: AC dan sistem AC. V). Pengertian sistem, piranti, komponen dan kaitan satu sama lain dalam teknik elektro. VI). Dasar elektronika VII). Dasar semikonduktor dan komponen semikonduktor
VIII). UTS.
Lanjutan Rincian
IX). Pengenalan pesawat-pesawa pesawat-pesawatt elektronika X). Pengantar analisa jaringan XI). Dasar-dasar teknik dijital XII). Komponen-komponen Komponen-komponen dijital dan dasar-dasar analisis rangkaian dijital XIII). Dasar-dasar rangkaian dijital XIV). Sejarah dan dasar teknik mikroprosesor mikroprosesor XV). Dasar teknik mikroprosesor dan pemrogramannya pemrogramannya XVI). Dasar penguat Op-Amp XVII). UAS
Daftar Pustaka Sumber Utama:& Richard C.Dorf: Ralph J.Smith C i r cu i ts , D evi e vi c e s, a n d S ys te m s ,
John Wiley & Sons,1995. F o u n d a ttii o n o f J.R.Cogdell: E l ectr e ctr i c a l E n g i n e e err i n g , Prentice Hall,1995. David E.Johnson, Johny
R.Johnson, : E le l e c t r i c C iirr cJohn u i t A nL.Hilburn al y s i s , Prentice Hall.
Referensi/Pengayaan Referensi: 1. P.H .H.. S Sm mal ale, e, , Te llee c o m m u n i c a t i o n S y s t e m I , Pitman Publishing Limited, London, 1978. 2. R.M .Maargunadi, P en en g a n t a r U m u m E l e k t r o t e k n i k , P.T.Dian Rakyat, Jakarta, 1986. 3. All llen en,, Mot Motte ters rshe head ad,, E l e c t r o n i c D e v i c e s an an d C i r c u i t s , a n i n t r o d u c t i o n , Prentice-Hall of India, New Delhi, 1976. 4. Enja Enjang ng A. Jua Juanda nda dan Ja Jaja ja K Kust ustija, ija, P en en g a n t a r E l e k t r o T e ek knik , JPTEFPTK-IKIP, Bandung, 1994. 5. A.P .P.. Ma Malv lvin ino, o, E l ec ec t r o n i c s P r i c i p l e s , Mc.Graw-Hill Mc.Graw -Hill Company Company,, London, 1985 6. Bri Brian an Moore Moore aand nd Joh John nD Dona onaghy ghy,, O p e r a t io , io n a l A m p l i f i e r C i r c u i t s Heinemann, London, 1986. 7. Ar Archie chie W W.Culp,J .Culp,Jrr (T (Terjem erjemahan: ahan: Ir Ir.. Darwin Sitomp Sitompul ul M.Eng M.Eng), ), P r i n s i p - p r i n s i p K o n v e r s i E n e rrg g i , Penerbit Erlangga, Jakarta, 1985. - Jurnal 1. IEEE, Telecommunication Transactions. - Internet Dosen dapat dihubungi melalui: Alamat rumah dan telpon: Jl. Suryalaya IX No.31 Bandung 40265T.7310350 Alamatt e-mail:
[email protected] Alama
Apperse Appersepsi psi
ARUS SEARAH (DC) (Arus dan Tegangan Listrik)
Arus Listrik 19 coulomb Q = 1.6 X 10 – 19
i = dq/dt 1 A = 1 coulomb/det Tipe Ti pe Besar Arus: Stasiun Pembangkit : 1000 100 0 A Starter Mobil
: 100 A
Lampu Dop
: 1A
Radio Mini
: 10 mA
Jam Tangan
: 1 mikroA
DC Circuits: Review • Current: The rate of flow of electric charge past a point in a circuit – Measured in amperes (A) – 1 A = 1 C/s = 6.25 1018 electrons per second – Current direction taken as direction positive direction positive charges flow – Analogou Analogous s to volume flow rate (volume/unit time) time) of water in a pipe
• Voltage: Electrical potential energy per unit charge – Measured in volts (V): 1 V = 1 J/C – Ground Ground is is the 0 V reference point point – Analogou Analogous s to water pressure
• Resistance: Restriction to charge flow – Measured in ohms (W) – Analogou Analogous s to obstacles that restrict water water flow
A Simple DC Circuit V
V
• Resistors have a constant resistance over a broad range of voltages and currents – Then V IR law) IR with R = constant (Ohm’s law) • Power = rate energy is delivered to the resistor = 2 rate energy is dissipated by the P IV I 2 R V resistor R
Hukum Ohm
Bahwa I V (hub. linier) Ditulis: V = I R Atau
: R = V/I
Bahan Non-Linier i
0
v
Power (Daya): energi yang diberikan pada elektron tiap satuan waktu
P = v dq/dt =vI 1 watt = 1 volt x 1 A Contoh Daya : Generator : 300 MW Radiator
: 1000 W
Lampu Senter : 6 W Jam Tangan : 10 mikroW
Ideal Voltage and Current Sources • An ideal voltage source is a source of voltage with zero internal resistance (a perfect battery) – Supply the same voltage regardless of the amount of current drawn from it
• An ideal current source a constant regardless of what load supplies it is connected to current – Has infinite internal resistance – Transistors can be represented by ideal current sources
(Introductory Electronics, Electronics, Simpson, 2nd Ed.) Ed.)
Ideal Voltage and Current Sources • Load resistance R L connected to terminals of a real current source: – Larger current is through the smaller resistance
(Introductory Electronics, Electronics, Simpson, 2nd Ed.) Ed.)
• Current sources can always be converted to voltage sources – Terminals A’B’ A’B’ act electrically exactly like terminals AB
(Introductory Electronics, Electronics, Simpson, 2nd Ed.) Ed.)
Contoh: Berapa arus yang mengalir pada resistor???
a. VA = 6 V VB = 2 V VAB = 4 V I=4A b. ……………
Komponen Rangkaian DC a) baterai, b) resistor dan c) kabel penghubung
Resistor standar: Toleransi 10% 10,12,15,18,22,27,33,39, 47,56,68,dan 82
Short circuit (hubung singkat: V =0 (R = 0) Open Circuit (hubung terbuka): I = 0 (R = ~)
Hukum Kirchhoff I.
Total arus pada suatu titik cabang = 0 I=0
II.
Total penur penurunan unan tegangan pada rangkaian tertutup = 0 V=0
Resistor Seri dan Paralel Seri:
masing-masing dilewati arus
yang sama RT = R1 +R2 +R3
Paralel: masing-masing mendapat tegangan yang sama 1/RT = 1/R1 + 1/R2 + 1/R3
Penyederhanaan Penyederhana an Rangkaian
Bagaimana Bagaiman a resistansi tergantung pada dimensi R = l/A (tergantung dimensi)
Seri?
Paralel?
Pembagi Potensial (Potential Divider)
Voltage Divider • Voltage divider: Circuit that produces a predictable fraction of the input voltage as the output voltage • Schematic: R 1
(Student Manual for The Art of Electronics,, Hayes and Horowitz, Electronics 2nd Ed.) Ed.)
R2
V in
• Current (same everywhere) is: I R R 1 2 • Output voltage (V out) is then given by: V out
IR2
R2
V in
R1 R2
Voltage Divider • Easier way to calculate
V out:
Notice the voltage drops
are proportional to the resistances R1
– For example, if R R1 = R2 then V out = V in / 2 – Another example: example: If R R1 = 4 W and R2 = 6 W, then V = (0.6)V out
R2
in
• No Now wa att ttac ach h a ““lo load ad”” res resis isto tor r R R L across the output: R1
R2
R1
R L
=
R2 R R L
– You can model R1 and R L as one resistor (parallel
combination), then calculate
V out
for this new voltage divider
Tugas : Tentukan besarnya v3 !
Pembagi tegangan terbebani
Penyederhanaan Penyederhana an Rangkaian
Voltage Division v1
i s R1
v2
and
i s R2
Applying KVL to the loop, v s
v1 v2 i s ( R1 R2 )
i s
and
v s R1 R2
Combining these yields the basic voltage division formula:
v1
v s
R1 R1 R2
v2
v s
R2 R1 R2
Voltage Division (cont.) Using the derived equations with the indicated values, v1
v2
10 V 10 V
8 k W 8.00 V 8 k W 2 k W 2 k W
2.00 V
8 k W 2 k W Design Note: Voltage division only applies when both resistors are carrying the same current.
Teorema Thevenin Jika suatu kumpulan rangkaian sumber potensial dan resistor dihubungkan dengan dua terminal keluaran, maka rangkaian tersebut dapat digantikan dengan sebuah rangkaian seri dari sebuah sumber potensial rangkaian terbuka dan sebuah resistor
Thevenin and Norton Equivalent Circuits
Find the Thevenin Equivalent Voltage
Problem: Find the Thevenin
equivalent voltage at the output.
Solution: •
Known Information topologyand andGiven Data: Circuit values in figure.
•
Unknowns: Thevenin equivalent voltage v TH.
•
TH Approach: Voltage source v TH
is defined as the output voltage with no load.
Assumptions: None. • Analysis: Next slide… slide…
•
Thevenin’s Theorem Theorem • Thevenin’s Theore rem m: Any combination of voltage sources and resistors with 2 terminals is electrically equivalent to an ideal voltage source in series with a single resistor RTh
(Introductory Electronics, Electronics, Simpson, 2nd Ed.) Ed.)
V Th
– Terminals A’B’ A’B’ electrica electrically lly equival equivalent ent to terminals AB AB
• Thevenin equivalent
V Th
V Th V (open circuit )
and RTh given by: RTh
(output voltage with no load attached) I (short
V (open circuit ) I (short circuit )
circuit) = current when the output is
shorted directly to ground
Thevenin’s Theorem Theorem • The Theven venin’ in’s s ttheo heorem rem app applie lied d tto o a vol voltag tage ed divi ivider der:: R1
R2
V Th
V out IR2
RTh
I (short circuit ) V in R1
R2 V in R1 R2
V (open circuit )
V Th
I (short circuit )
I (short circuit )
R1 R2 1
R
R2
• Thevenin equivalent circuit: RTh
(Introductory Electronics, Electronics, Simpson, 2nd Ed.) Ed.) V Th
(a load resistance R L can then be attached between terminals A’ and B’, in series with RTh)
R2 – Note that RTh = R1 R
• Imagine mentally shorting out the voltage source • Then R1 is in parallel with R2
• Only works for constant (independent) voltage sources (batteries)
Contoh: Dengan menggunakan teorema Thevenin, hitung besarnya arus I 2 pada rangkaian di bawah ini
Vo/c = 5,455 V RP = 5,455 k Ohm
I2 = 0,397 mA
Menurut Thevenin
Example Problem #1.9
(The Art of Electronics, Electronics, Horowitz and Hill, 2nd Ed.) Ed.)
Solution (details given in class): (a) 15 V (b) 10 V (c)
V Th
= 15 V, RTh = 5k
(d) 10 V
(e)
P L
= 0.01 W, P R2 = 0.01 W, P R1 = 0.04 W
Example Problem #1.7 Solution (details given in class): 1 – –V V source: 0.667 V 10k – –10k 10k voltage divider: 0.4 V
THÉVENIN EQUIVALENT CIRCUITS
The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.
Vt Voc We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.
The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.
Voc
Z t I sc
Vt
I n I sc
I sc
CURRENT DEVIDER (Pembagi Arus) io = iI (G2/G1+G2)
Current Division (cont.) Using the derived equations with the indicated values, i1
5 ma 5 ma
i 2
3 k W 3.00 mA 2 k W 3 k W 2 k W
2.00 mA
2 k W 3 k W
Design Note: Current division only applies when the same voltage appears across both resistors.
Current Division i s
i1 i2
where i1
v s R1
and
i2
v s R2
Combining and solving for v , s
v s
i s
1
R1 R2
i s i s R1 || R2 1 1 R1 R2
R1
R2
Combining these yields the basic current division formula: i1
i s
R2 R1 R2
i 2 i s
R1 R1 R 2
and
Norton’s Theorem Theorem • Norton’s Theorem: Any combination of voltage sources and resistors with 2 terminals is electrically equivalent to an ideal current source in parallel with a single resistor I N
(Introductory Electronics, Electronics, Simpson, 2nd Ed.) Ed.)
R N
– Terminals A’B’ A’B’ electrica electrically lly equival equivalent ent to terminals AB AB
• Norton equivalent I N and R N given by: R N
RTh
V (open circuit ) I (short circuit )
(same as Thevenin equivalent resistance)
I N
V (open circuit ) R N
Norton’s Theorem Theorem • No Nort rton on’s ’s the theor orem em ap appl plie ied d to a volt voltag age e divi divide der: r: R1
R2
I N
R N
V in R1 V (open circuit )
R1 R2 R1 R2
I N
• Norton equivalent circuit: (Introductory Electronics, Electronics, Simpson, 2nd Ed.) Ed.) N I
R N
(a load resistance R L can then be attached between terminals A’ and B’, in parallel in parallel with R N )
– The Norton equivalent circuit is just as good as the
Thevenin equivalent circuit, and vice versa
TRANSMISI LISTRIK
Konstruksi Transformator
Persamaan gelombang sinus: y
a s in ( t )
2 gelombang dengan amplitudo berbeda tetapi berfase awal sama
y
4 s in t
y
2 s in t
2 gelombang dengan amplitudo sama tetapi berfase awal berbeda
y
y
4 s in t
4 s in ( t / 4)
2 gelombang dengan amplitudo sama tetapi berfase awal berbeda
y 1
4 s in ( t / 2)
y 2
4 s in ( t / 2)
Bagaimana jika y1 + y2 ?
Superposisi dua gelombang
Diagram Phasor:
Rangkaian resistor murni
v V s in t i I s in t i dan v sefase
Rangkaian Kapasitor Murni
v V s in t V c os t i Xcc X Xcc 1 / C X i mendahului v sebesar 90o
Rangkaian Induktor Murni
i I s in t v I X XL L cos t XL X L L o
v mendahului i sebesar 90
Rangkaian RC untuk Tapis Lolos Rendah vi = vR + vC
Sudut fase diambil dari input ke output
Perbandingan keluaran dan masukan:
V o / V i
1 / 1 ( R C )
untuk o t g
(
2
1 / RC ( saat X C R ) / o )
Bagaimana V o / V i untuk o
i )
i i )
i i i )
o
V o / V i
1/ 1 (
/ o )
2
)
o
Grafik vo/vi terhadap frekuensi (di kertas semilog: linier-logaritmik
Frekuensi 3 dB Besarnya Penguatan (Gain) biasa dinyatakan dengan dB, dengan definisi:
dB 20 log10 V o / V i
Untuk : o
1 / 2 saat X C
V o / V i
diperoleh Gain o
3dB
R
atau
disebut frekuensi 3dB
Rangkaian RC untuk Tapis Lolos Tinggi vi = vR + vC
Sudut fase diambil dari input ke output
Perbandingan keluaran dan masukan:
V o / V i
1 / 1 (1 / R C ) 2
Bagaimana V o / V i untuk
untuk o
o
1 / RC ( saat X C R )
o
i )
i i )
o
t g
( V o / V i
/ ) 1/ 1 (
i i i ) o
/ )
o
2
Grafik vo/vi terhadap frekuensi
Steady-State Sinusoidal A Analysis nalysis
Sinusoidal Currents and Voltages Phasors Complex Impedances Circuit Analysis with Phasors&Complex Impedances Power in AC Circuits Thevenin and Norton Equivalent Circuits Balanced Three-Phase Circuits
SINUSOIDAL CURRENTS AND VOLTAGES V m is the peak value ω is the angular frequency in radians per second θ is θ is the phase angle T is T is the period
Frequency
f
1 T
Angular frequency
2 T
2 f
s in z cos z 90
Root-Mean-Square Root-Mean-Squa re Values V r m s
P a vg
1
T
T
1
2 0
2 rm s
V R
v t dt I rm s
T 2
T 0 i t dt
P avg I R 2 rm s
RMS Value ofV am Sinusoid V rm s
2
The rms value for a sinusoid is the peak value divided by the square root of two. This is not true for other periodic waveforms such as square waves or triangular waves.
Phasor Definition Time function : v1 t V 1 cosωt θ 1 Phasor : V1 V 1θ 1
Adding Sinusoids Using Phasors Step 1: Determine the phasor for each term. Step 2: Add the phasors using complex arithmetic. Step 3: Convert the sum to polar form. Step 4: Write the result as a time function.
Using Phasors to Add 20 cos t 45 v1 t Sinusoids
v 2 t 10 cos t 60
V1 20 45
V2
10
30
Vs
V1
20 45
V2
14 .14
10
j14 .14
30
8.660
23 .06 j19 .14
29 .97 39 .7
j5
v s t 29 .97 cos t 39 .7
Sinusoids can be visualized as the realaxis projection of vectors rotating in the complex plane. The phasor for a sinusoid is a snapshot of the corresponding rotating vector at t = 0.
Phase Relationships To determine determine phase relationships from a phasor diagram, consider the phasors to rotate at a counterclockwise. Then when standing fixed point, if V1 arrives first followed by V2 2 1 leads after a rotation of of θ θ , , we say that V V by θ θ .. Alternatively Alternatively,, we could say say that V2 lags θ .. (Usually θ as V1 by θ (Usually,, we take θ as the smaller angle between the two phasors.)
To determine determine phase relationships relationships between sinusoids from their plots versus time, find the shortest time interval t between positive p peaks of the two waveforms. Then, the phase angle is θ = (t p /T /T )) × 360°. If the peak of v of v 1(t ) occurs first, we say that v 1(t ) leads v 2(t ) or that v 2(t ) lags v 1(t ). ).
COMPLEX IMPEDANCES V L j L I L
Z L j L L90
V L Z L I L
VC
Z C I C
Z C 1 1 90 C j 1 C j C C
V R RI R
Kirchhoff’s Laws in Phasor
Form We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path. The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.
Circuit Analysis Using Phasors and Impedances 1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)
2. Replace inductances by their complex =
L jωL impedances Z jωL..complex Replaceimpedances capacitances by their Z C = 1 /(jωC) /(jωC).. Resistances have impedances equal to their resistances. using any o off the techniques techniques 3. Analyze the circuit using studied earlier in Chapter 2, performing the calculations with complex arithmetic.
AC Power Calculations P V rm s I r m s cos PF co coss
v i
Q V r m s I rm s s in
apparent po pow wer V rm s I rm s
P Q V rm s I rm s 2
P I R 2 rm s
Q
2 rm s
I
X
2
2
P
Q
2 Rr m s
V
R 2 X r m s
V
X
Maximum Average Power Transfer If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance. If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the
magnitude of the Thévenin impedance.
BALANCED THREE-PHASE CIRCUITS Much of the power used by business and industry is supplied by three-phase distribution systems. Plant engineers need to be familiar with three-phase power.
Phase Sequence Three-phase sources can have either a positive or negative phase sequence. The direction of rotation of certain three-phase motors can be reversed by changing the phase sequence.
Wye –Wye Connection
Three-phase sources and loads can be connected either in a wye configuration or in a delta configuration. The key to understanding the various threephase configurations is a careful examination of the wye– wye –wye circuit.
wye wye circuit.
P a vg p t 3V Y rm sI Lrm s cos
Q3
V Y I L
2
s in 3V Y r m s I Lr m s s in
Z
Y
3 Z
