lecture 2d
TRANSCRIPT
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Lecture #2 By Yohandri 1
Rangkaian Pembag i
Tegangan dan A rus
Voltage and Current Divider Circuit
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Kompetensi Dasar
Mahasiswa dapat menganalisis rangkaian
pembagi tegangan dan pembebanan,rangkaian pembagi arus serta rangkaian
setara Thevenin dan Norton
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1. Series And Paral lel Circu its
Circuits consisting of just one battery and one load
resistance are very simple to analyze, but they are notoften found in practical applications. Usually, we find
circuits where more than two components are connected
together. There are two basic ways in which to connect
more than two circuit components: series and parallel.
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1.1. Resis to rs in series
The first principle to understand about series circuits is
that the amount of current is the same through any
component in the circuit. This is because there is only
one path for electrons to flow in a series circuit, and
because free electrons flow through conductors.
Two resistors, R1 and R2, connected in series have
voltage drop V = I(R1 + R2). That is, they have a
combined resistance Rs given by their sum:
This generalizes for n series resistors to
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OR
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1.2. Resis to rs in paral lel
Two resistors, R1 and R2, connected in parallel have
voltage drop V = I.Rp, where
21
111
RRRP
This generalizes for n parallel resistors to
21
21
RR
RRR
P
R1
R2
There are many paths for electrons
to flow, but only one voltage across all components
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OR
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2.1. Voltage Divider
2. Voltage Devider and Load ing (Pembagi tegangan dan pembebanan)
A voltage divider consists of two
resistances R1 and R2 connected in
series across a supply voltage Vs. The
supply voltage is divided up between
the two resistances to give an output
voltage Vo which is the voltage across
R2. This depends on the size of R2
relative to R1
VsRR
RVo
21
2
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2.2. Pembebanan
Apabila keluaran suatu rangkaian dibebani maka
pada keluaran rangkaian tersebut dapat terjadi
penurunan tegangan atau jatuh tegangan, peristiwa
ini disebut pembebanan
E
R2
R1
RL
VsRR
RVo
21
2
Tegangan keluaran tanpa RL
Tegangan keluaran dengan RL
)//( 21 LRRR
E
I
)//.( 2 LOL RRIV
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Contoh
E
R2
R1 1K
1K Vo
12VE
R2
R1
RL
1k
1k
12V
ERR
RVo
21
2
)//( 21 LRRR
EI
)//.( 2 LOL RRIV
Tegangan keluaran tanpa RL
VVo 61211
1
Tegangan keluaran dengan RL
Misal RL = 1K
mAI 8)1//1(1
12
VVOL
4)1//1.(8
Vo = 6 4 = 2V
Jatuh tagangan
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3. Current Devider
A parallel circuit is often called a current divider for its
ability to proportion or divide the total current intofractional parts
E R2R1
I
I1 I2 E = I RT RT = R1//R2
I1 = VR1/ R1
For parallel circuit VR1 = VR2 = E
Current through R1
IR
RI
T
1
1 21
21
RR
RRR
T
IRR
RR
RI
2
1
1
21
1
1
I
RR
RI
21
2
1
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4. Rangkaian Setara Thevenin dan Norton
Rangkaian setara sangat penting dalam elektronika, dengan
rangkaian setara dapat dilakukan pengukuran pada
masukan dan keluaran suatu alat atau rangkaian tanpaperlu tahu bentuk rangkaian di dalamnya.
Ada dua bentuk dasar rangkaian setara yaitu
Rangkaian setara Thevenin
Rangkaian setara Norton
Rangkaian setara Thevenin : Menggunakan sumber
tegangan tetap, yakni suatu sumber tegangan ideal
dengan tegangan keluaran yang tidak berubah,
berapapun besarnya arus yang diambil darinya.
Rangkaian setara Norton : Menggunakan sumber arus
tetap, yang dapat menghasilkan arus tetap, berapapun
besarnya hambatan yang dipasang pada keluarannya
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4.1. Rangakaian Setara Thevenin
Setiap rangkaian dengan dua ujung atau gerbang tunggal,
dapat digantikan dengan suatu sumber tegangan tetap
atau suatu gaya gerak listrik (ggl) dan suatu hambatan
seri dengan ggl tersebut
Ri Ro
Ei EoVo
ViRo
ETHVo
Thevenin's Theorem
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Analisis Rangkaian Thevenin
E
R2
R1
Vo
21
21
RR
RRR
TH
ERR
RE
TH
21
2
RTH dapat dihitung dengan menggantiE dengan hubungan singkat, sehingga
Dan ETH dapat dihitung dengan sistem
pembagi tegangan
Contoh :
Tentukan rangkaian setara
UntukR1 = 1 K
R2 = 1 K
Bagaimana jika
R1 = 100 R2 = 100
Amati jika pada keluaranya
dari model 1 dan 2 diambil
arus 10 mA
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R1
E
R2 R3 Vo
Latihan
Ro
ETHVo
Diketahui : R1 = 1 K
R2 = 2 K
R3 = 1 K
E = 12V
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4.2. Rangakaian Setara Norton
Rangkaian setara ini terdiri dari sumber arus tetap INparalel dengan suatu hambatan R
O
RoIN
Ro
ETH
Bila ujung kelauarn Thevenin dan Norton sama-sama
dihubung singkatkan maka arus yang mengalir pada
keluaran akan bernilai sama
N
THI
Ro
RIos
Iosadalah arus keluaran jika
dihubungkan singkat
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R1
E
R2 R4
R3
Example
R1 = 1K
R2 = 2K
R3
= 1K
R4 = 1K
E = 12V