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Pobierz
Chapter 4
Electrical
Fundamentals
DC Circuits and Resistance
Glossary
Alternating current
— A flow of
charged particles through a conductor,
first in one direction, then in the other
direction.
Ampere
— A measure of flow of
charged particles per unit time. One
ampere represents one coulomb of charge
flowing past a point in one second.
Atom
— The smallest particle of mat-
ter that makes up an element. Consists of
protons and neutrons in the central area
called the nucleus, with electrons sur-
rounding this central region.
Coulomb
— A unit of measure of a
quantity of electrically charged particles.
One coulomb is equal to 6.25 × 10
18
elec-
trons.
Direct current
— A flow of charged
particles through a conductor in one direc-
tion only.
EMF
— Electromotive Force is the
term used to define the force of attraction
between two points of different charge
potential. Also called voltage.
Energy
— Capability of doing work. It
is usually measured in electrical terms as
the number of watts of power consumed
during a specific period of time, such as
watt-seconds or kilowatt-hours.
Joule
— Measure of a quantity of
energy. One joule is defined as one new-
ton (a measure of force) acting over a dis-
tance of one meter.
Ohm
— Unit of resistance. One ohm is
defined as the resistance that will allow
one ampere of current when one volt of
EMF is impressed across the resistance.
Power
— Power is the rate at which
work is done. One watt of power is equal
to one volt of EMF, causing a current of
one ampere through a resistor.
Volt
— A measure of electromotive force.
Introduction
The DC Circuits and Resistance section
of this chapter was written by Roger Tay-
lor, K9ALD.
The atom is the primary building block
of the universe. The main parts of the atom
include protons, electrons and neutrons.
Protons have a positive electrical charge,
electrons a negative charge and neutrons
have no electrical charge. All atoms are
electrically neutral, so they have the same
number of electrons as protons. If an atom
loses electrons, so it has more protons than
electrons, it has a net positive charge. If an
atom gains electrons, so it has more elec-
trons than protons, it has a negative
charge. Particles with a positive or nega-
tive charge are called ions. Free electrons
are also called ions, because they have a
negative charge.
When there are a surplus number of
positive ions in one location and a surplus
number of negative ions (or electrons) in
another location, there is an attractive
force between the two collections of par-
ticles. That force tries to pull the collec-
tions together. This attraction is called
electromotive force, or EMF.
If there is no path (conductor) to allow
electric charge to flow between the
two locations, the charges cannot move
together and neutralize one another. If a
conductor is provided, then electric cur-
rent (usually electrons) will flow through
the conductor.
Electrons move from the negative to the
positive side of the voltage, or EMF
source.
Conventional current
has the
opposite direction, from positive to nega-
tive. This comes from an arbitrary deci-
sion made by Benjamin Franklin in the
18th century. The conventional current
direction is important in establishing the
proper polarity sign for many electronics
calculations. Conventional current is used
in much of the technical literature. The
arrows in semiconductor schematic sym-
bols point in the direction of conventional
current, for example.
To measure the quantities of charge,
current and force, certain definitions have
been adopted. Charge is measured in
cou-
lombs
. One coulomb is equal to 6.25 × 10
18
electrons (or protons). Charge flow is
measured in
amperes
. One ampere repre-
sents one coulomb of charge flowing past
a point in one second. Electromotive force
is measured in
volts
. One volt is defined as
the potential force (electrical) between
two points for which one ampere of cur-
rent will do one
joule
(measure of energy)
of work flowing from one point to another.
(A joule of work per second represents a
power of one watt.)
Electrical Fundamentals
4.1
Voltage can be generated in a variety of
ways. Chemicals with certain characteris-
tics can be combined to form a battery.
Mechanical motion such as friction (static
electricity, lightning) and rotating con-
ductors in a magnetic field (generators)
can also produce voltage.
Any conductor between points at dif-
ferent voltages will allow current to pass
between the points. No conductor is per-
fect or lossless, however, at least not at
normal temperatures. Charged particles
such as electrons resist being moved and it
requires energy to move them. The amount
of resistance to current is measured in
ohms
.
Then:
Table 4.1
Relative Resistivity of Metals
Resistivity Compared
E = 0.150 A × 20000 Ω = 3000 V
Material
to Copper
When 150 V is applied to a circuit, the
current is measured at 2.5 A. What is the
resistance of the circuit? In this case R is
t
he unknown, so we will use equation 1:
Aluminum (pure)
1.60
Brass
3.7-4.90
Cadmium
4.40
Chromium
1.80
Copper (hard-drawn)
1.03
E
150
V
Copper (annealed)
1.00
R
=
=
=
60
Ω
Gold
1.40
I
2.5
A
Iron (pure)
5.68
No conversion was necessary because
the voltage and current were given in volts
and amperes.
How much current will flow if 250 V is
applied to a 5000-Ω resistor? Since I is
u
nknown,
Lead
12.80
Nickel
5.10
Phosphor bronze
2.8-5.40
Silver
0.94
Steel
7.6-12.70
Tin
6.70
Zinc
3.40
OHM’S LAW
One ohm is defined as the amount of
resistance that allows one ampere of cur-
rent to flow between two points that have
a potential difference of one volt. Thus,
we get Ohm’s Law, which is:
E
250
V
I
=
=
=
0.05
A
R
5000
Ω
It is more convenient to express the
current in mA, and 0.05 A × 1000 mA / A
= 50 mA.
tional area, the one with the larger area
will have the lower resistance.
RESISTANCE OF WIRES
The problem of determining the resis-
tance of a round wire of given diameter
and length—or its converse, finding a suit-
able size and length of wire to provide a
desired amount of resistance—can easily
be solved with the help of the copper wire
table given in the
Component Data and
References
chapter. This table gives the
resistance, in ohms per 1000 ft, of each
standard wire size. For example, suppose
you need a resistance of 3.5 Ω, and some
#28 wire is on hand. The wire table in the
Component Data and References
chap-
ter shows that #28 wire has a resistance of
66.17
E
RESISTANCE AND
CONDUCTANCE
Suppose we have two conductors of the
same size and shape, but of different
materials. The amount of current that will
flow when a given EMF is applied will
vary with the resistance of the material.
The lower the resistance, the greater the
current for a given EMF. The
resistivity
of
a material is the resistance, in ohms, of a
cube of the material measuring one centi-
meter on each edge. One of the best
conductors is copper, and in making
resistance calculations it is frequently
convenient to compare the resistance of
the material under consideration with that
of a copper conductor of the same size and
shape.
Table 4.1
gives the ratio of the
resistivity of various conductors to the
resistivity of copper.
The longer the physical path, the higher
the resistance of that conductor. For direct
current and low-frequency alternating
currents (up to a few thousand hertz) the
resistance is inversely proportional to the
cross-sectional area of the path the cur-
rent must travel; that is, given two con-
ductors of the same material and having
the same length, but differing in cross-sec-
R =
(1)
I
where:
R = resistance in ohms,
E = potential or EMF in volts and
I = current in amperes.
Transposing the equation gives the
other common expressions of Ohm’s Law
as:
E = I × R
(2)
a
nd
/ 1000 ft. Since the desired resis-
tance is 3.5 Ω, the required wire length is:
Ω
E
I =
(3)
R
All three forms of the equation are used
often in radio work. You must remember
that the quantities are in volts, ohms and
amperes; other units cannot be used in the
equations without first being converted.
For example, if the current is in milliam-
peres you must first change it to the
equivalent fraction of an ampere before
substituting the value into the equations.
The following examples illustrate the
use of Ohm’s Law. The current through a
20000-Ω resistance is 150 mA. See
Fig
4.1
. What is the voltage? To find voltage,
use equation 2 (E = I × R). Convert the
current from milliamperes to amperes.
Divide by 1000 mA / A (or multiply by
10
-3
A / mA) to make this conversion.
(Notice the conversion factor of 1000 does
not limit the number of significant figures
in
t
he calculated answer.)
R
3.5
Ω
DESIRED
=
Length
=
R
66.17
Ω
WIRE
1000
ft
1000
ft
3.5
Ω
×
1000
ft
=
=
53
ft
(4)
66.17
Ω
As another example, suppose that the
resistance of wire in a circuit must not
exceed 0.05 Ω and that the length of wire
required for making the connections totals
1
4 ft. Then:
R
R
0.05
Ω
WIRE
MAXIMUM
〈
=
(5)
1000
ft
Length
14.0
ft
Ω
1000
ft
−
3
=
3.57
×
10
×
ft
1000
ft
R
WIRE
〈
3.57
Ω
1000
ft
1000
ft
150
mA
Find the value of R
WIRE
/ 1000 ft that is
less than the calculated value. The wire
table shows that #15 is the smallest size
I
=
=
0.150
A
mA
Fig 4.1 — A simple circuit consisting of
a battery and a resistor.
1000
A
4.2
Chapter 4
having a resistance less than this value.
(The resistance of #15 wire is given as
3.1810
Fig 4.2 — Examples
of various resistors.
At the top left is a
small 10-W
wirewound resistor.
A single in-line
package (SIP) of
resistors is at the top
right. At the top
center is a small PC-
board-mount
variable resistor. A
tiny surface-mount
(chip) resistor is also
shown at the top.
Below the variable
resistor is a 1-W
carbon compo-sition
resistor and then a ½-W composition unit. The
dog-bone-shaped resistors at the bottom are ½-W
and ¼-W film resistors. The ¼-inch-ruled graph
paper background provides a size comparison.
The inset photo shows the chip resistor with a
penny for size comparison.
/ 1000 ft.) Select any wire size
larger than this for the connections in your
circuit, to ensure that the total wire resis-
tance will be less than 0.05 Ω.
When the wire in question is not made
of copper, the resistance values in the wire
table should be multiplied by the ratios
shown in Table 4.1 to obtain the resulting
resistance. If the wire in the first example
were made from nickel instead of copper,
t
he length required for 3.5
Ω
Ω
would be:
R
DESIRED
Length
=
R
WIRE
(6)
1000
ft
3.5
Ω
=
66.17
Ω
×
5.1
1000
ft
3.5
Ω
×
1000
ft
=
66.17
Ω
×
5.1
3500
ft
Length
=
=
10.37
ft
337.5
In addition, special material or construc-
tion techniques may be used to provide
temperature compensation, so the value
does not change (or changes in a precise
manner) as the resistor temperature
changes. There is more information about
the electrical characteristics of real resis-
tors in the
Real-World Component
Characteristics
chapter.
rent through a resistance causes the
conductor to become heated; the higher
the resistance and the larger the current,
the greater the amount of heat developed.
Resistors intended for carrying large cur-
rents must be physically large so the heat
can be radiated quickly to the surrounding
air. If the resistor does not dissipate the
heat quickly, it may get hot enough to melt
or burn.
The amount of heat a resistor can safely
dissipate depends on the material, surface
area and design. Typical carbon resistors
used in amateur electronics (
1
/
8
to 2-W
resistors) depend primarily on the surface
area of the case, with some heat also being
carried off through the connecting leads.
Wirewound resistors are usually used for
higher power levels. Some have finned
cases for better convection cooling and/or
metal cases for better conductive cooling.
In some circuits, the resistor value may
be critical. In this case, precision resistors
are used. These are typically wirewound,
or carbon-film devices whose values are
carefully controlled during manufacture.
TEMPERATURE EFFECTS
The resistance of a conductor changes
with its temperature. The resistance of
practically every metallic conductor
increases with increasing temperature.
Carbon, however, acts in the opposite
way; its resistance decreases when its tem-
perature rises. It is seldom necessary to
consider temperature in making resistance
calculations for amateur work. The tem-
perature effect is important when it is nec-
essary to maintain a constant resistance
under all conditions, however. Special
materials that have little or no change in
resistance over a wide temperature range
are used in that case.
CONDUCTANCE
The reciprocal of resistance (1/R) is
conductance
. It is usually represented by
the symbol G. A circuit having high con-
ductance has low resistance, and vice
versa. In radio work, the term is used
chiefly in connection with electron-tube
and field-effect transistor characteristics.
The unit of conductance is the siemens,
abbreviated S. A resistance of 1 Ω has a
conductance of 1 S, a resistance of 1000
RESISTORS
A package of material exhibiting a cer-
tain amount of resistance, made up into a
single unit is called a resistor. Different
resistors having the same resistance value
may be considerably different in physical
size and construction (see
Fig 4.2
). Cur-
Ω
has a conductance of 0.001 S, and so on. A
unit frequently used in connection with
electron devices is the µS or one millionth
of a siemens. It is the conductance of a
1-M
Ω
resistance.
Series and Parallel Resistances
Very few actual electric circuits are as
simple as Fig 4.1. Commonly, resistances
are found connected in a variety of ways.
The two fundamental methods of connect-
ing resistances are shown in
Fig 4.3
. In
part A, the current flows from the source
of EMF (in the direction shown by the ar-
row) down through the first resistance, R1,
then through the second, R2 and then back
to the source. These resistors are con-
nected in series. The current everywhere
in the circuit has the same value.
In part B, the current flows to the com-
mon connection point at the top of the two
resistors and then divides, one part of it
flowing through R1 and the other through
R2. At the lower connection point these
two currents again combine; the total is
the same as the current into the upper com-
mon connection. In this case, the two
resistors are connected in parallel.
Electrical Fundamentals
4.3
RESISTORS IN PARALLEL
In a circuit with resistances in parallel,
the total resistance is less than that of the
lowest resistance value present. This is
because the total current is always greater
than the current in any individual resistor.
The formula for finding the total resistance
of resistances in parallel is:
Fig 4.4 — An example of resistors in
parallel. See text for calculations.
Fig 4.5 — An example of resistors in
series. See text for calculations.
1
R
=
1
1
1
1
Ohm’s Law, as shown below. The current
through R1 is I1, I2 is the current through
R2 and I3 is the current through R3.
For convenience, we can use resistance
in kΩ, which gives current in milliam-
p
eres.
(7)
+
+
+
+
...
R1
R2
R3
R4
where the dots indicate that any number of
resistors can be combined by the same
method. For only two resistances in paral-
lel (a very common case) the formula
b
ecomes:
connected to a source of EMF as shown in
Fig 4.5
. The EMF is 250 V, R1 is 5.00 k
,
R2 is 20.0 kΩ and R3 is 8.00 kΩ. The total
resistance is then
Ω
E
250
V
I1
=
=
=
50.0
mA
R1
5.00
kΩ
R
TOTAL
=
R1
+
R2
+
R3
R1
×
R2
R
=
E
250
V
(8)
I2
=
=
=
12.5
mA
R1
+
R2
=
R = 33.0 kΩ.
T
he current in the circuit is then
R
5.00
kΩ
+
20.0
kΩ
+
8.00
kΩ
R2
20.0
kΩ
resistor is con-
nected in parallel with one of 1200 Ω, what
i
s the total resistance?
Example: If a 500-
Ω
E
250
V
I3
=
=
=
31.2
mA
R3
8.00
kΩ
E
250
V
R1
×
R2
500
Ω
×
1200
Ω
Notice that the branch currents are
inversely proportional to the resistances.
The 20000-
I
=
=
=
7.58
mA
R
=
=
R
33.0
kΩ
R1
+
R2
500
Ω
+
1200
Ω
resistor has a value four times
larger than the 5000-Ω resistor, and has a
current one quarter as large. If a resistor
has a value twice as large as another, it will
have half as much current through it when
they are connected in parallel.
The total circuit current is:
I3
Ω
(We need not carry calculations beyond
three significant figures; often, two will
suffice because the accuracy of measure-
ments is seldom better than a few percent.)
2
600000
Ω
R
=
=
353
Ω
1700
Ω
KIRCHHOFF’S FIRST LAW
(KIRCHHOFF’S CURRENT LAW)
Suppose three resistors (5.00 kΩ,
20.0 k
KIRCHHOFF’S SECOND LAW
(KIRCHHOFF’S VOLTAGE LAW)
Ohm’s Law applies in any portion of a
circuit as well as to the circuit as a whole.
Although the current is the same in all
three of the resistances in the example of
Fig 4.5, the total voltage divides between
them. The voltage appearing across each
resistor (the voltage drop) can be found
from Ohm’s Law.
Example: If the voltage across R1 is
called E1, that across R2 is called E2 and
t
hat across R3 is called E3, then
I
TOTAL
=
I1
+
I2
+
(9)
) are connected in
parallel as shown in
Fig 4.4
. The same
EMF, 250 V, is applied to all three resis-
tors. The current in each can be found from
Ω
and 8.00 k
Ω
I
TOTAL
=
50.0
mA
+
12.5
mA
+
31.2
mA
I
TOTAL
=
This example illustrates Kirchhoff’s
Current Law: The current flowing into a
node or branching point is equal to the sum
of the individual currents leaving the node
or branching point. The total resistance of
t
he circuit is therefore:
93.7
mA
E
=
IR1
=
0.00758
A
×
5000
Ω
=
37.9
V
E
250
V
R
=
=
=
2.67
kΩ
I
93.7
mA
E
=
IR2
=
0.00758
A
×
20000
Ω
=
152
V
You can verify this calculation by com-
bining the three resistor values in parallel,
using equation 7.
=
Notice here that the voltage drop across
each resistor is directly proportional to the
resistance. The 20000-
E
IR3
=
0.00758
A
×
8000
Ω
=
60.6
V
resistor value is
four times larger than the 5000-Ω resistor,
and the voltage drop across the 20000-Ω
resistor is four times larger. A resistor that
has a value twice as large as another will
have twice the voltage drop across it when
they are connected in series.
Kirchhoff’s Voltage Law accurately de-
scribes the situation in the circuit: The sum
of the voltages in a closed current loop is
zero. The resistors are power sinks, while
the battery is a power source. It is common
to assign a + sign to power sources and a
Ω
RESISTORS IN SERIES
When a circuit has a number of resis-
tances connected in series, the total
resistance of the circuit is the sum of the
individual resistances. If these are num-
bered R1, R2, R3 and so on, then:
R
TOTAL
=
R1
+
R2
+
R3
+
R4
...
(10)
where the dots indicate that as many resis-
tors as necessary may be added.
Example: Suppose that three resistors are
Fig 4.3 — Resistors connected in series
at A, and in parallel at B.
4.4
Chapter 4
This resistance in series with R1 forms
a simple series circuit, as shown in Fig
4.6B. The total resistance in the circuit is:
E
– sign to power sinks. This means the volt-
ages across the resistors have the opposite
sign from the battery voltage. Adding all
the voltages yields zero. In the case of a
single voltage source, algebraic manipula-
tion implies that the sum of the individual
voltage drops in the circuit must be equal
t
o the applied voltage.
E3
I
=
(12)
R1
+
R2
and the voltage between terminals A and
B (E
AB
) is:
R
=
R1
+
R
=
5.00
kΩ
+
5.71
kΩ
TOTAL
EQ
R
TOTAL
=
10.71
kΩ
E
AB
=
I
×
R2
(13)
T
he current is:
E
TOTAL
=
E1
+
E2
+
(11)
By substituting the first equation into
the second, we can find a simplified
expression for E
AB
:
E
250
V
E
TOTAL
=
37.9
V
+
152
V
+
60.6
V
I
=
=
=
23.3
mA
R
10.71
kΩ
E
TOTAL
=
(Remember the significant figures rule
for addition.)
In problems such as this, when the cur-
rent is small enough to be expressed in
milliamperes, considerable time and
trouble can be saved if the resistance is
expressed in kilohms rather than in ohms.
When the resistance in kilohms is substi-
tuted directly in Ohm’s Law, the current
will be milliamperes, if the EMF is in volts.
250
V
The voltage drops across R1 and R
EQ
are:
R2
E
AB
=
×
E
(14)
R1
+
R2
E1
=
I
×
R1
=
23.3
mA
×
5.00
kΩ
=
117
V
E2
=
I
×
R
=
23.3
mA
×
5.71
kΩ
=
133
V
EQ
with sufficient accuracy. These two volt-
age drops total 250 V, as described by
Kirchhoff’s Current Law. E2 appears
across both R2 and R3 so,
RESISTORS IN SERIES-PARALLEL
A circuit may have resistances both
in parallel and in series, as shown in
Fig 4.6A
. The method for analyzing such
a circuit is as follows: Consider R2 and R3
to be the equivalent of a single resistor,
R
EQ
whose value is equal to R2 and R3 in
p
arallel.
E2
133
V
I2
=
=
=
6.65
mA
R2
20.0
kΩ
E3
133
V
I3
=
=
=
16.6
mA
R3
8.00
kΩ
where:
I2 = current through R2 and
I3 = current through R3.
R2
×
R3
20000
Ω
×
8000
Ω
R
EQ
=
=
R2
+
R3
20000
Ω
+
8000
Ω
The sum of I2 and I3 is equal to 23.3 mA,
conforming to Kirchhoff’s Voltage Law.
8
2
1.60
×
10
Ω
=
THEVENIN’S THEOREM
Thevenin’s Theorem is a useful tool
for simplifying electrical networks.
Thevenin’s Theorem states that any two-
terminal network of resistors and voltage
or current sources can be replaced by a
single voltage source and a series resistor.
Such a transformation can simplify the
calculation of current through a parallel
branch. Thevenin’s Theorem can be
readily applied to the circuit of Fig 4.6A,
to find the current through R3.
In this example, R1 and R2 form a volt-
age divider circuit, with R3 as the load
(
Fig 4.7A
). The current drawn by the load
(R3) is simply the voltage across R3,
divided by its resistance. Unfortunately,
the value of R2 affects the voltage across
R3, just as the presence of R3 affects the
potential appearing across R2. Some
means of separating the two is needed;
hence the Thevenin-equivalent circuit.
The voltage of the Thevenin-equivalent
battery is the open-circuit voltage, mea-
sured when there is no current from either
terminal A or B. Without a load connected
between A and B, the total current through
the circuit is (from Ohm’s Law):
28000
Ω
R
EQ
=
5710
Ω
=
5.71
kΩ
Fig 4.7 — Equivalent circuits for the
circuit shown in Fig 4.6. A shows the
load resistor (R3) looking into the
circuit. B shows the Thevenin-
equivalent circuit, with a resistor and a
voltage source in series. C shows the
Norton-equivalent circuit, with a
resistor and current source in parallel.
Fig 4.6 — At A, an example of resistors
in series-parallel. The equivalent circuit
is shown at B. See text for calculations.
Electrical Fundamentals
4.5
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