dc circuits and methods of circuits analysis circuits elements: voltage source current source...
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DC circuits and methods of circuits analysis
• Circuits elements:• Voltage source• Current source• Resistors• Capacitors • Inductors
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Voltage source - V [V]
• Ideal sourceConstant output voltage, internal resistance equals to zero
• Real sourceOutput voltage depends on various conditions. Dependence may be linear (battery) on non-linear
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Current source - I [A]
• Ideal sourceConstant output current, internal resistance equals to infinity
• Real sourceOutput current depends on various conditions. Dependence may be linear on non-linear (Usually electronic sources)
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Resistance - R []
• Coductance G=1/R [S]
• Ideal resistorlinear R = const.V= I . R
• Real resistornon-linear(electric bulb, PN junction)
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Resistance (2)
• Resistors in seriesR = R1 + R2
• Resistors in parallelR = R1 // R2 = (R1 . R2) / (R1 + R2)
• Voltage dividerU2 = U . R2 /(R1 + R2)potential divider (‘pot’)
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Passive electronic parts
• Resistors feature electrical resistivity RR
dimensioning according maximal dissipation power (loses) Pmax
• Capacitors feature capacity CC dimensioning according maximal granted voltage Vmax
• Inductors feature inductivity LLdimensioning according maximal granted current Imax
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Resistors• Feature: resistivity• r = R = const.• nonreversible el. energy transfer to heat• • • Data: R [Ω], P [W]• Description: Ω → J, R 4,7 Ω → 4R7• kΩ → k 68 kΩ → 68k• MΩ → M 2.2 MΩ → 2M2• 0,15 MΩ
→ M15• 47k/0,125W 3R3/ 5W
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ResistorsResistors color codings
First strip is near to edge than last If tolerance is ±20 %, the 4. strip miss
colornumber
tolerance
Blacká 0
Brown 1 ± 1 %
Red 2 ± 2 %
Orange 3
Yelow 4
Green 5 ± 0,5 %
Blue 6 ± 0,25 %
violet 7 ± 0,1 %
grey 8
white 9
gold -1 ± 5 %
silver -2 ± 10 %
no color ± 20 %
Meaning
Strip 4 strips 5strips
1 first digit first digit
2 second digit second digit
3 exponent 10x third digit
4 tolerance exponent 10x
5 tolerance
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Resistors
Material
• Carbon – non stable, temperature dependent
• Metalised - stable, precise
• Wired more power dissipation > 5W
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Resistors
Potentiometer variable resistor
Potentiometr adjustable by hand Potentiometer adjustable by tool
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Resistors
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Capacitors
• Part: Capacitor, condenser• Feature: capacity
t
dtiC
v0
.1
dynamic definition
symbol
c = C = const.
Accumulator of the energy in electrostatic field
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Capacitors
tivCq .
2.2
1VCW
static definition
power definition
unit: 1 F (Farrad) dimension: [A.s/V]
For calculation should be used SI system only! :
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Capacitors
Description: • pF → J, R 4,7 pF → 4R7• 103 pF → k , n 68 000 pF → 68k• 106 pF → M 3,3 µF → 3M3
• 109 pF → G 200 µF → 200M
Number code: number, number, exponent in pF
eg. : 474 → 470 000pF → 470k → M47 ±20%
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Capacitors
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InductorsPart: Inductor, coilFeature: inductivity
dt
diLu dynamic definition
l = L = konst.
Accumulator of the energy in electrostatic field
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Inductors
2.2
1ILW
static definition
power definition
unit: 1 H (Henry) dimension: [V.s/A]
For calculation should be used SI system only! :
ILN ..
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Inductors
Details for instalation and ordering
L [H], IMAX [A]
Lower units 1 µH = 10-3 mH = 10-6 H
-------------------
It use in electronic not very often.
See next semestr
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Ohm’s and Kirchhoff’s laws
• Ohm’s law I = U / R• 1st Kirchhoff’s law (KCL) I = 0
At any node of a network, at every instant of time, the algebraic sum of the currents at the node is zero
• 2nd Kirchhoff’s law (KVL) U = 0 The algebraic sum of the voltages across all the components around any loop of circuits is zero
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Nodal analysis (for most circuits the best way)
• Uses 1st K. law – Chose reference node– Label all other voltage nodes– Eliminate nodes with fixed voltage by source
of emf– At each node apply 1st K. law– Solve the equations
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Mesh analysis
• Uses 2nd K. law– Find independent meshs– Eliminate meshs with fixed current source– Across each mesh apply 2nd K. law– Solve the equations
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Thevenin equialent circuitfor linear circuit
As far as any load connected across its output terminals is concerned, a linear circuits consisting of voltage sources, current sources and resistances is equivalent to an ideal voltage source VT in series with a resistance RT. The value of the voltage source is equal to the open circuit voltage of the linear circuit. The resistance which would be measured between the output terminals if the load were removed and all sources were replaced by their internal resistances.
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Norton equialent circuitfor linear circuit
As far as any load connected across its output terminals is concerned, a linear circuits consisting of voltage sources, current sources and resistances is equivalent to an ideal current source IN in parallel with a resistance RN. The value of the current source is equal to the short circuit voltage of the linear circuit. The value of the resistance is equal to the resistance measured between the output terminals if the load were removed and all sources were replaced by their internal resistances.
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Principle of superposition
• The principle of superposition is that, in a linear network, the contribution of each source to the output voltage or current can be worked out independently of all other sources, and the various contribution then added together to give the net output voltage or current.
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Example
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Methods of electrical circuits analysis:
• Node Voltage Method Σii = 0 , ΣIi = 0
• Mesh Current Method Σvi = 0 , ΣVi = 0 • Thevenin and Norton Eq. Cirtuits• Principle of Superposition
• --- and other 15 methods
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Topology and Number of Lineary Independent Equations
No. of elements p p No. of voltage sources zzvv
No. of nodes u u No. of current sources zzii
R1 R 3 R 2
V1 V 2
3
I 1I 2
I3
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• No of elements p = 5 p = 5 No of voltage sources zzvv = 2 = 2
• No. of nodes u = 4 u = 4 No of current sources zzii = 0 = 0
• No of independent nodes XXii = u – 1 = u – 1 - z- zuu = 4 – = 4 – 11 - - 22 = 1 = 1
• No of independent meshes XXii = p – u + 1 = p – u + 1 – z– zii = 5 – 4 + 1 = 2= 5 – 4 + 1 = 2
R1 R 3 R 2
V1 V U2
V 3
I 1I 2
I3
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Node Voltage Analysis Method
1. Select a reference node (usually ground). All other node voltages will be referenced to this node.
2. Define remaining n-1 node voltages as the independent variables.
3. Apply KCL at each of the n-1 nodes, expressing each current in terms of the adjacent node voltages
4. Solve the linear system of n-1 equations in n-1 unknowns
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R1 R 3 R 2
V1 V 2
V 3
03
3
2
23
1
13
R
V
R
VV
R
VV
321
2
2
1
1
3 111RRR
RV
RV
V
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Mesh Current Analysis Method
1. Define each mesh current consistently. We shall define each current clockwise, for convenience
2. Apply KVL around each mesh, expressing each voltage in terms of one or more mesh currents
3. Solve the resulting linear system of equations with mesh currents as the independent variables
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________________
R1 R 3 R 2
V1 V 2
V 3
I 1I 2
I3
0121311 VIIRIR
0222123 VIRIIR
123131 VIRIRR
223213 VIRRIR