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ECE 3336
Introduction to Circuits & Electronics
Note Set #2
Circuit Elements, Ohm’s Law,
Kirchhoff’s Laws
Spring 2015,
TUE&TH 5:30-7:00 pm
Dr. Wanda Wosik
1
Basic Elements of the Circuits
Voltage sources
Current sources
Resistors
Independent and Dependent
•
•
Independent sources are real
Dependent sources represent behavior
of other systems/elements (later)
Ohm’s Law
Current flows through the resistor
Voltage builds-up on the resistor terminals.
2
The 5 Basic Circuit Elements
There are 5 basic circuit elements:
1.
Voltage sources
2.
Current sources
3.
Resistors
4.
Inductors (later)
5.
Capacitors (later)
3
Voltage Sources
• A voltage source is a two-terminal circuit element that
maintains a voltage V across its terminals.
• The value of the voltage is the defining characteristic of a
voltage source. Location in the circuit and polarity of this
source is important.
• Any value of the current can go through the voltage
source, in any direction. The current can also be zero.
The voltage source does not “care about” current. It
“cares” only about voltage.
Real voltage
source has a
resistance in
series
4
Voltage Sources - Battery
Figure
2.32, 2.33
Voltage Source and Current Direction
i
i
D
D
A battery supplies voltage, which allows for current (see conventions
for the current vs. positive charges or electrons) to flow through a
diode D to a lamp. When the battery is discharging the light becomes
deemed and then goes off.
6
Voltage Sources
Voltage sources:
1.
Independent voltage sources (DC and AC)
2.
Dependent voltage sources, of which there are 2 forms:
i.
Voltage-dependent voltage sources
+
ii.
Current-dependent voltage sources
7
Voltage Sources
Symbol for Independent Sources
Independent AC
voltage sources
Battery
vS=
#[V]
+
-
Independent
voltage
source
The schematic symbol can be labeled either with a variable,
like vS, or a value and units.
8
Ideal Voltage Sources
Various Representations of an
Electrical System
Practical Voltage
Source
Using the Voltmeter to Measure
Voltage.
Nigel P. Cook
Electronics: A Complete Course, 2e
11 Inc.
Copyright ©2004 by Pearson Education,
Upper Saddle River, New Jersey 07458
All rights reserved.
Measurement of Voltage
Dependent Voltage Sources
• Voltage-dependent voltage
sources; µ[V/V] dimensionless
vS=
m vX
+
-
• Current-dependent voltage
sources; r[V/A] has a
dimension
vS=
r iX
+
-
Voltagedependent
voltage
source
Currentdependent
voltage
source
13
Current Sources
• A current source is a two-terminal circuit element
that maintains a current through its terminals,
independently on the circuit connection.
Ideal current sources
• The value and sign of the current source is given.
• Any voltage can be across the current
source, in either polarity. It can also be zero.
The current source does not “care about” voltage.
It “cares” only about current.
Real current sources
Just add R in parallel
NEVER assume that you know current
in the voltage source!
Calculate the voltage value and find its
direction!
14
Current Sources
Current sources:
1. Independent current sources
2.
Dependent current sources:
i. Voltage-dependent current sources
ii. Current-dependent current sources
15
Practical Current
Source
16
Measurement of Current
Dependent Current Sources
i.
ii.
Voltage-dependent
current sources;
coefficient g has
dimension [A/V]
Current-dependent
current sources;
coefficient b is
dimensionless [A/A]
The values for these coefficients are
always shown without units.
iS=
g vX
Voltagedependent
current
source
iS=
b iX
Currentdependent
current
source
18
Voltage and Current Polarities
•
Importance of reference polarities of currents and
voltages.
•
Notice that the schematic symbols for the voltage
sources and current sources indicate these polarities.
•
The voltage sources have a “+” and a “–” to show the
voltage reference polarity.
•
•
The current sources have an arrow
current reference polarity.
to show the
19
Units
• We will show units for the values of the coefficients r and g in a
problem description not on the schematic.
• Variables should not have units.
• Values of variables must have the units.
•Examples:
vX = 120 V
iQ = 35 A
pabs = 24.5 kW
pdel = vQ(13 A)
pabs = vXiX
•Examples of missing
units:
vX = 1.5
pdel = 25 iQ
iX = 15
20
Resistors
• A resistor is a two terminal element
that has a constant ratio of the voltage
to the current through its terminals.
+
iR
vR
-
R=vR/iR
R=vR/iR
• It’s unit is Ohm or Ω=V/A
• The resistor does not have polarity.
• IMPORTANT: use Ohm’s Law only
on resistors. It does not hold for
sources
21
Microscopic View of Ohm’s Law
Ohm’s Law
Drift velocity
Conductivity
22
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html#c1
Resistor Polarities
•
There is no corresponding polarity to a resistor. The resistor is
symmetrical (like a pipe of specified x-section).
•
•
However, direction REALLY matters.
The voltage and current is in the passive sign convention – then
power (vi) dissipated by the resistor is
Current is treated as a flow of
positive charges so it flows
from higher to lower potential
i.e. from “+” to “-”.
R
+
iR
-
v
RX=
#[W]
+
vX
iX
-
In the active sign
vX
convention we would RX = iX
have ix<0 and power
delivered <0
RESISTANCE
(=dissipation)
CANNOT BE <0
23
Resistors, Resistance, and Ohm’s Law
I(A)
• Resistance of RESISTORS is always positive.
• It can have some nonlinear behavior (ex. for larger currents)
R[W] =
R=V/I
l[cm]
r[Wcm]
2
A[cm ]
V(V)
• Resistance, as a ratio of voltage to current, can be negative in some
devices/conditions:
I(A)
• Devices with negative resistance provide positive power.
• This can be seen in dependent sources (various elements).
V(V)
24
The Resistance and Resistor
2-8
Ohmmeter and Measurement of
Resistance
Kirchhoff’s Laws
• Extremely useful laws for currents and voltages
in circuits:
• Kirchhoff’s Current Law (KCL)
• Kirchhoff’s Voltage Law (KVL)
• Here: connections of elements by wires will be
very IMPORTANT.
• Wires have ≈ ∞ (high) conductivity (s) so R≈0[Ω]
1
s [S / cm] =
r[Wcm]
27
Nodes
Important for KCL
• A node is defined as a point where two or more
components are connected.
branches
• We connect components with wires, which do not
cause voltage drop (Rwire=0 Ω).
hmm, really?
28
Find the Nodes
• How many nodes are there
in this circuit?
• Remember that wires have
R≈0Ω so V=0V
RC
vA
RD
+
-
5? no because there are
RE
2 pairs of the same
nodes
so the answer is 3
RF
iB
29
Closed Loops
Important for KVL
•
•
•
A closed loop is a completely closed contour in the circuit.
It starts and ends at the same point.
It can, but does not need, to contain elements!
•
It can also jump across open space i.e. will not follow components i.e. elements
do not need to be connected.
30
Find Closed Loops
Example circuit
•
Several closed loops are
possible.
•
The total number of simple
closed loops in this circuit is 13.
RC
vA
RD
+
+
-
vX
-
iB
RE
RF
31
Closed Loop #1
RC
• Loop #1 shown red.
vA
RD
+
+
-
vX
-
iB
RE
RF
32
Closed Loop #2
RC
• Loop #2 shown red.
vA
RD
+
+
-
vX
-
iB
RE
RF
33
Closed Loop #3
RC
• Loop #3 shown red.
• Jump across the
voltage labeled vX.
vA
RD
+
+
-
vX
-
iB
RE
RF
34
Closed Loop #4
RC
• Loop #4 shown red.
• The same jump across the
voltage labeled vX.
• The loop also crossed the
current source.
vA
RD
+
+
-
vX
-
Remember that a current source
can have a voltage across it.
Do not assume that you know
this voltage but calculate its
value and direction
iB
RE
RF
35
A Not-Closed Loop
RC
• Not a closed contour
i.e. not a closed loop
vA
RD
+
+
-
vX
-
iB
RE
RF
36
Kirchhoff’s Current Law (KCL)
•
The algebraic (or signed i.e with directions) summation of
currents through a closed surface (such as a node) must equal
zero.
n
å ik = 0
It means that there is no charge build-up there.
k=1
37
Demonstration of
Kirchhoff’s Current
Law
38
Kirchhoff’s Current Law (KCL)
Expand the KCL concept to a part of the circuit i.e. bigger node -> super node
n
å ik = 0
k=1
Here currents in this KCL:
IS1, I3 and I5
It means that charge does not build up at this big node so charge is
conserved. We may be not interested what happens with other
currents? Hmm?
39
Current Polarities in KCL
In KCL we have to assign a sign to each
reference current polarity that we will use
(and do not change it from assignment to
assignment!).
We can choose:
+ sign for all currents entering the node
- sign for all currents leaving the node
or
- sign for all currents entering the node
+ sign for all currents leaving the node
These conventions are exactly opposite
but equivalent in KCL.
n
å ik = 0
k=1
40
KCL in the Example Circuit
We will use a convention (w/o
flip-floping):
RC
+ for currents leaving the node
– for those that are entering.
iA
vA
•For this circuit with our notation
in KCL, we have the following
equation:
iC
RD
+
iD
RE
-iA + iC - iD + iE - iB = 0
iE
iB
iB
RF
41
KCL More Intuitive Approach
• Currents entering the node
(closed surface) must be
equal all currents leaving this
node.
iA + iD + iB = iC + iE
RC
iC
iA
vA
RD
+
iD
-
• This is the same as before.
RE
-iA + iC - iD + iE - iB = 0
It is the consequence of KCL
iE
iB
iB
RF
n
å ik = 0
k=1
42
Kirchhoff’s Voltage Law (KVL)
The algebraic (or signed i.e. directions defined for
voltages) summation of voltages around a closed
loop must equal zero.
It results from energy conservation
within a closed loop
n
å vk = 0
k=1
-v1+v2=0
43
KVL and Voltage Polarities
In any closed loop we must designate a sign to each
reference voltage polarity. We can (but will not) choose:
+ if the voltage shows a rise (- to +)
or
- if the voltage shows a drop (+ to -)
as we circulate around the closed loop.
Once the convention is set DO NOT change it from assignment to
assignment. v2
v4
-
+
v1
+
+
n
-
å vk = 0
+
+
v3
k=1
v5
-
44
KVL – an Example
We will use the following convention:
• positive sign assigned for a voltage drop
• negative sign assigned to a voltage rise.
-
v2
+
+
v4
v1
+
-
å vk = 0
k=1
-
+
v3
n
+
v5
-
KVL in each loop will give total
three equations
Loop 1: v1-v2+v3=0
Loop 2: -v3+v4+ v5=0
Loop 3: v1-v2 + v4+v5=0
45
KVL an Example (from the loops ex.)
• positive sign assigned for a
voltage drop
• negative sign assigned to a
voltage rise.
RC
vA
• KVL, when starting at the
bottom, will give the following
equation:
+
Entering
Positive
RD
+
-
vX
RE
-
-vA + vX - vE + vF = 0
Entering
Negative
iB
+
-
vE
+
RF
vF
See mnemonics
-
46
KVL More Intuitive Notation
• Voltage drops must equal the
voltage rises
RC
vX + vF = vA + vE
vA
• Earlier we had
RD
+
+
-
vX
-vA + vX - vE + vF = 0
RE
-
+
-
These are the same equations.
n
å
vk
It is the consequence of
of KVL
k=1
You can use either method.
=0
vE
+
RF
iB
vF
-
47
Equations to Describe the Circuit
• To solve a circuit we must determine unknown currents, voltages and, if
required, also power.
• For DC conditions, KVL, KCL and Ohm’s Laws will be enough. We will
add techniques to make solutions easy and systematic.
• Number of equations must be equal # of unknowns.
48
Ohm’s Law, KVL and KCL
Find the voltage vX and the current iX. Use the following steps.
+
v4
-
for KVL
R4=
20[W]
vS1=
3[V]
+
-
iX
R3=
100[W]
i3 - for Ohm’s Law
R4=
+ v4 20[W]
+
vX vS1=
3[V]
+
-
-
v X = i3 R3
iX
R3=
100[W]
Polarities give
v4 = -i X R4
R4=
+ v4 20[W]
+
vX
i3
-
v4 and i3 (defined)
vS1=
3[V]
+
-
+
iX
R3=
100[W]
vX
i3
-
Solve
-vS1 + v4 + v X = 0, or
i X + i3 = 0, or
-3[V] + v4 + v X = 0.
i3 = -i X .
49
Ohm’s Law, KCL and KVL
Results for vx and ix
R4=
+ v4 20[W]
• Solve the equations
-3[V] - iX 20[W] - iX 100[W] = 0, or
-3[V]
iX =
= -25[mA].
120[W]
vS1=
3[V]
+
-
We have substituted into our KVL equation from
other equations:
-vS1 + v4 + v X = 0, or
v X = - (-25[mA])100[W] = 2.5[V]. -3[V]+ v4 + v X = 0.
+
iX
R3=
100[W]
vX
i3
-
v4 = -iX R4 , and
v X = i3 R3 .
iX + i3 = 0, or
i3 = -iX50.