First Class Course

Assistant Professor Dr. Laith Abdullah Mohammed

Email: dr.laith@uotechnology.edu.iq
Website: www.uotechnology.edu.iq/dep-production/laith/
►Requirements and Grading:
First term Exam: 15%
Second term Exam: 15%
Final Exam: 50%
Homworks, Quizess and Class attendance: 10%
Laboratory: 10%

►Course Materials:
1 - Edward Hughes, [Electrical and Electronic Technology], 11th edition, Prentice
Hall, 2012
2 - Austin Hughes , [ Electric Motors and Drives: Fundamentals, Types and
Applications], 3rd edition, Newnes, 2005
3- Richard Crowder, [ Electric Drives and Electromechanical systems ], Butterworth
Hainemann, 2006

►Lectures:

Lectures are available on department’s website:

www.uotechnology.edu.iq/dep-production
  Definition of Basic Electrical Quantities
 Kirchoff’s Law in Voltage and Current
►Course topics: 

Analysis of Electric Circuits
Determination of Equivalent Resistance
 Series/Parallel Circuits
 Delta-Star Transformation
These topics are for 30  Thevenin’s Theorem
weeks.  Norton’s Theorem
 Super-position Theorem
 Maximum Power Transfer
 Complex Numbers
 Response of inductive Circuit
 Response of Capacitive Circuit
 Response of R-L-C Circuit
 Principle of Electromechanical Energy Conversion
 Principle of operation of DC Motors
 Equivalent Circuit of DC Motors
 Types of DC Motors
 Speed –Torque curve Characteristic of DC Motors
 Speed Control of DC Motor
 DC Drives
 Principle of Operation of AC Motors
 Types of AC Motors
 Equivalent Circuit of AC Motors
 Speed –Torque Curve Characteristics of AC Motors
 Speed Control of AC Motors
 AC Drives
 Stepper Motors
 Theory of servo Motors
 Response of servo Motors
Definition of Basic Electrical Quantities
Ideal Voltage source:
It provides a prescribed voltage across its terminals irrespective of the current flowing through it. The
amount of current supplied by the source is determined by the circuit connected to it.

Various representations of an electrical system

 Ideal Current source:
It provides a prescribed current to any circuit connected to it. The voltage
generated by the source is determined by the circuit connected to it.

An electrical network is a collection of elements through which current flows.

The following definitions introduce some important elements of a network.

Elements of an electrical network

Branch:
A branch is any portion of a circuit with two terminals connected to it. A branch may consist of
one or more circuit elements . In practice, any circuit element with two terminals connected to it is
a branch.
Node
A node is the junction of two or more branches . In effect, any connection that can be
accomplished by soldering various terminals together is a node. It is very important to identify
nodes properly in the analysis of electrical networks.

Loop
is any closed connection of branches.
The fundamental electric quantity is charge, and the smallest amount of charge that exists is the
charge carried by an electron, equal to:
qe = −1.602 × 10−19 C
the amount of charge associated with an electron is rather small.
This, of course, has to do with the size of the unit we use to measure charge, the coulomb (C)
Current consists of the flow of very large numbers of charge particles.

The other charge-carrying particle in an atom, the proton, is assigned a plus sign and the same
magnitude. The charge of a proton is
qp = +1.602 × 10−19 C
Electrons and protons are often referred to as elementary charges.

Electric current is defined as the time rate of change of charge passing through a predetermined area
(Typically, this area is the cross-sectional area of a metal wire).

imagine ∆ q units of charge flowing through the cross-sectional area A in ∆t units of time.
The resulting current i is then given by:
i= ∆q/ ∆t C/s
The units of current are called amperes, where 1 ampere (A) = 1 coulomb/second (C/s).
In order for current to flow, there must exist a closed circuit.

The current i flowing from the battery to the lightbulb is equal to

the current flowing from the lightbulb to the battery.
In other words, no current (and therefore no charge) is “lost”
around the closed circuit.
This principle was observed by the German scientist
G. R. Kirchhoff and is now known as
Kirchhoff’s current law (KCL).
Kirchhoff’s current law states that because charge cannot be
created but must be conserved, the sum of the currents at
a node must equal zero. A simple electric circuit

The resulting expression for node 1

of the circuit is:
−i + i1 + i2 + i3 = 0 (assuming currents entering a node as being
negative)

Note that if we had assumed that currents entering the node

were positive, the result would not have changed.
Example: Determine the unknown currents in the circuit of Figure below.
IS = 5A I1 = 2A I2 = −3A I3 = 1.5A
Find: I0 and I4.
Solution:

Two nodes are clearly shown in Figure as node a and node b; the third node in the
circuit is the reference (ground) node. Apply KCL at each of the three nodes.
At node a:
I0 + I1 + I2 = 0
I0 + 2 − 3 = 0
∴ I0 = 1A
Note that the three currents are all defined as flowing away from the node, but one
of the currents has a negative value (i.e., it is actually flowing toward the node).

At node b:
IS − I3 − I4 = 0
5 − 1.5 − I4 = 0
∴ I4 = 3.5A
Note that the current from the battery is defined in a direction opposite to that of
the other two currents (i.e., toward the node instead of away from the node). Thus,
in applying KCL, we have used opposite signs for the first and the latter two
currents.
At the reference node: If we use the same convention (positive value for currents
entering the node and negative value for currents exiting the node), we obtain the
following equations:
−IS + I3 + I4 = 0
−5 + 1.5 + I4 = 0
∴ I4 = 3.5A
VOLTAGE AND KIRCHHOFF’S VOLTAGE LAW (KVL) Lecture .2
Charge moving in an electric circuit gives rise to a current.
Naturally, it must take some work, or energy, for the charge to
move between two points in a circuit, say, from point a to point b.

The total work per unit charge associated with the motion of
charge between two points is called voltage.
Thus, the units of voltage are those of energy per unit charge; they
have been called volts.

The voltage, or potential difference, between two points in a circuit indicates the energy required
to move charge from one point to the other converting the potential energy within the voltage
source to electric power. the direction, or polarity, of the voltage is closely tied to whether energy
is being dissipated or generated in the process.
The principle underlying KVL is that no energy is lost or created in an electric circuit, the sum of all
voltages associated with sources must equal the sum of the load voltages, so that the net voltage
around a closed circuit is zero.

where the vn are the individual voltages around the closed circuit.
Example:
Figure below depicts the battery pack in the electric car. the series connection of 31 (12-V)
batteries that make up the battery supply for the electric vehicle.
Find: Expression relating battery and electric motor drive voltages.

Solution:

the electric drive is supplied by a 31×12 = 372-V battery pack.

Lecture No.2 2
Example: Determine the unknown voltage V2 of the circuit shown in figure.
VS2 = 12 V V1 = 6 V V3 = 1 V

Solution: Applying KVL around the simple loop, we write

VS2 − V1 − V2 − V3 = 0
V2 = VS2 − V1 − V3 = 12 − 6 − 1 = 5 V

Example: Determine the unknown voltages V1 and V4 in the circuit of Figure below.
VS1 = 12 V VS2 = −4 V V2 = 2 V V3 = 6 V V5 = 12 V

Solution: To determine the unknown voltages, we apply KVL

clockwise around each of the three meshes:
VS1 − V1 − V2 − V3 = 0
V2 − VS2 + V4 = 0
V3 − V4 − V5 = 0
Next, we substitute numerical values:
12 − V1 − 2 − 6 = 0
V1 = 4 V
2 − (−4) + V4 = 0
V4 = −6 V
6 − (−6) − V5 = 0
V5 = 12 V

Lecture No.2 3
 voltage as work per unit charge
power is the work done per unit time

The electric power generated by an active element, or that dissipated or stored by

a passive element, is equal to the product of the voltage across the element and
the current flowing through it. P=VI units of power is (joules per second, or watts).
power is a signed quantity.

circuit A and the direction of the current through it indicate that the circuit is
doing work in moving charge from a lower potential to a higher potential.
circuit B is dissipating energy, because the direction of the current indicates that
charge is being displaced from a higher potential to a lower potential.
THE PASSIVE SIGN CONVENTION
1. Choose an arbitrary direction of current flow.
2. Label polarities of all active elements (voltage and current sources).
3. Assign polarities to all passive elements (resistors and other loads); for passive
elements, current always flows into the positive terminal.
4. Compute the power dissipated by each element according to the following rule:
If positive current flows into the positive terminal of an element, then
the power dissipated is positive (i.e., the element absorbs power);
if the current leaves the positive terminal of an element, then the power dissipated is
negative (i.e., the element delivers power).

Lecture No.2 4
Example: Apply the passive sign convention to the circuit of Figure
beside. The voltage drop across load 1 is 8 V, that across load 2 is 4 V;
the current in the circuit is 0.1 A.
Find: Power dissipated or generated by each element.

sign convention is independent of the current direction we choose.

We now apply the method twice to the same circuit. Following the passive sign convention, we first select an
arbitrary direction for the current in the circuit; the example will be repeated for both possible directions of
current flow to demonstrate that the methodology is sound.
1. Assume clockwise direction of current flow, as shown in Figure (a).
2. Label polarity of voltage source, as shown in Figure (a); since the arbitrarily chosen direction of the current is
consistent with the true polarity of the voltage source, the source voltage will be a positive quantity.
3. Assign polarity to each passive element, as shown in Figure (a).
4. Compute the power dissipated by each element: Since current flows from − to + through the battery, the
power dissipated by this element will be a negative quantity:
PB = −vB × i = −12 V × 0.1A= −1.2W
that is, the battery generates 1.2 watts (W). The power dissipated by the
two loads will be a positive quantity in both cases, since current flows
from plus to minus:
P1 = v1 × i = 8 V× 0.1A = 0.8W
P2 = v2 × i = 4 V× 0.1A = 0.4W

Lecture No.2 5
Next, we repeat the analysis, assuming counterclockwise current direction.
1. Assume counterclockwise direction of current flow, as shown in Figure (b).
2. Label polarity of voltage source, as shown in Figure (b); since the arbitrarily chosen direction
of the current is not consistent with the true polarity of the voltage source, the source voltage
will be a negative quantity.
3. Assign polarity to each passive element, as shown in Figure (b).
4. Compute the power dissipated by each element: Since current flows from plus to minus
through the battery, the power dissipated by this element will be a positive quantity;
however, the source voltage is a negative quantity:
PB = vB × i = (−12 V)(0.1A) = −1.2W
that is, the battery generates 1.2 W, as in the previous case.
The power dissipated by the two loads will be a positive quantity in both cases, since current
flows from plus to minus:
P1 = v1 × i = 8 V× 0.1A = 0.8W
P2 = v2 × i = 4 V× 0.1A = 0.4W

Power supplied always equals power dissipated

Lecture No.2 6
Example: For the circuit shown in Figure below, determine which components are absorbing
power and which are delivering power.

Solution:
By KCL (Kirchhoff’s Current Law), the current through element B is 5 A, to the right.
By KVL (Kirchhoff’s Voltage Law),
−va − 3 + 10 + 5 = 0
Therefore, the voltage across element A is
va = 12 V (positive at the top)
A supplies (12 V)(5 A) = 60W
B supplies (3 V)(5 A) = 15W
C absorbs (5 V)(5 A) = 25W
D absorbs (10 V)(3 A) = 30W
E absorbs (10 V)(2 A) = 20W
Total power supplied = 60W+ 15W= 75W
Total power absorbed = 25W+ 30W+ 20W= 75W
Total power supplied = Total power absorbed,

so conservation of power is satisfied

Lecture No.2 7
RESISTANCE AND OHM’S LAW Lecture .3

When electric current flows through a metal wire or through other circuit elements, it encounters a
certain amount of resistance, the magnitude of which depends on the electrical properties of the
material. An ideal resistor is a device that exhibits linear resistance properties according to Ohm’s
law, which states that V=IR
The value of the resistance R is measured in units of ohms (Ω ), where 1 Ω = 1 V/A
The resistance of a material depends on a property called resistivity, denoted by the symbol ρ;
the inverse of resistivity is called conductivity and is denoted by the symbol σ.

For a cylindrical resistance element, the resistance is proportional to the length of the sample l
and inversely proportional to its cross sectional area A and conductivity σ.
Define the conductance of a circuit element as the inverse of its resistance. The symbol used
to denote the conductance of an element is G, where
G = 1/R siemens (S)
where 1 S = 1 A/V
Thus, Ohm’s law can be restated in terms of conductance as
I = GV

Resistivity of common materials

at room temperature

For a resistor R, the power dissipated can be expressed, with Ohm’s law by
P = VI = I 2 R = V2 / R

Lecture No.3 2
Example: Determine the minimum resistor size that can be
connected to a given battery without exceeding the resistor’s ¼ W
power rating.
Resistor power rating = 0.25W.
Battery voltages: 1.5 and 3 V.

Solution: We first need to obtain an expression for resistor power dissipation as a function
of its resistance.
We know that P = VI
V = IR.
Thus, the power dissipated by any resistor is
PR = V × I = V × V/R = V2 / R

Since the maximum allowable power dissipation is 0.25W, we can write

V2/R ≤ 0.25, or R ≥ V2/0.25.
Thus, for a 1.5 V battery, the minimum size resistor will be
R = 1.52/0.25 = 9 Ω.
For a 3 V battery the minimum size resistor will be
R = 32/0.25 = 36 Ω.
Lecture No.3 3
Series Resistors and the Voltage Divider Rule:

Parallel and series circuits have a direct relationship with Kirchhoff’s laws.
These circuits form the basis of all network analysis.
For an example of a series circuit, refer to the circuit of Figure, where a
battery has been connected to resistors R1,R2, andR3.

By applying KVL, you can verify that the sum of the voltages across the
three resistors equals the voltage externally provided by the battery
1.5 V = v1 + v2 + v3
And since, according to Ohm’s law, the separate voltages can be expressed
by the relations
v1 = iR1
v2 = iR2
v3 = iR3
we can therefore write
1.5 V = i(R1 + R2 + R3)
To the battery, the three series resistors appear as a single equivalent resistance of value REQ,
where:
REQ = R1 + R2 + R3
The three resistors could thus be replaced by a single resistor of value REQ without changing the
amount of current required of the battery.

Lecture No.3 4
 we can write each of the voltages across the resistors as:

The general form of the voltage divider rule for a circuit with N series resistors and a voltage
source is

Lecture No.3 5
Example: Determine the voltage v3 in the circuit of Figure.
R1 = 10 Ω ;
R2 = 6 Ω ;
R3 = 8 Ω ;
VS = 3 V

Solution: figure indicates a reference direction for the current (dictated by the
polarity of the voltage source).
Following the passive sign convention, we label the polarities of the three resistors,
and apply KVL to determine that
VS − v1 − v2 − v3 = 0

The voltage divider rule:

Lecture No.3 6
Parallel Resistors and the Current Divider Rule Lecture .4

Two or more circuit elements are said to be in parallel if the elements share the same terminals.
From KVL, it follows that the elements will have the same voltage.

Kirchhoff’s current law (KCL) requires that the sum of the currents into, say, the top node of the
circuit be zero:
iS = i 1 + i 2 + i 3
But by using of Ohm’s law we may express each current as follows:

since, by definition, the same voltage v appears across each element. Kirchhoff’s current law may
then be restated as follows:
 Note that this equation can be also written in terms of a single equivalent resistance REQ

we can generalize this result to an arbitrary number of resistors connected in parallel by stating
that N resistors in parallel act as a single equivalent resistance REQ given by the expression

Lecture No.4 2
Current divider rule
Consider the three resistor circuit of Figure. From the expressions already derived from each of
the currents i1, i2, and i3, we can write

Current
Divider
Rule

Lecture No.4 3
 Example: Determine the current i1 in the circuit of Figure
Where: R1 = 10 Ω ; R2 = 2 Ω ; R3 = 20 Ω ;IS = 4 A.

Solution: Application of the current divider rule yields

Example: Determine the voltage v in the circuit of Figure

Solution: The circuit of Figure is neither a series nor a parallel circuit because
the following two conditions do not apply:
1. The current through all resistors is the same (series circuit condition).
2. The voltage across all resistors is the same (parallel circuit condition).
The circuit takes a much simpler appearance once it becomes evident that
the same voltage appears across both R2 and R3 and, therefore, that these
elements are in parallel. If these two resistors are replaced by a single
equivalent resistor , now the equivalent circuit is a simple series circuit, and
the voltage divider rule can be applied to determine that

Lecture No.4 4
For series circuits, we can apply Voltage divider rule:
𝑅𝑅𝑛𝑛
𝑣𝑣𝑛𝑛 = 𝑣𝑣
𝑅𝑅1 + 𝑅𝑅2 + ⋯ + 𝑅𝑅𝑛𝑛 𝑠𝑠

𝑅𝑅2 ∥ 𝑅𝑅3
𝑣𝑣 = 𝑣𝑣
𝑅𝑅1 + 𝑅𝑅2 ∥ 𝑅𝑅3 𝑠𝑠

For series circuits, current is found by:

𝑣𝑣𝑠𝑠
𝑖𝑖 =
𝑅𝑅𝐸𝐸𝐸𝐸

𝑣𝑣𝑠𝑠
𝑖𝑖 =
𝑅𝑅1 + 𝑅𝑅2 ∥ 𝑅𝑅3

For parallel circuits:

1
𝑅𝑅2 ∥ 𝑅𝑅3 =
1/𝑅𝑅2 + 1/𝑅𝑅3

Lecture No.4 5
Example: Consider the circuit of Figure,
1. without resistor R3. Calculate the value of the voltage v
if the source voltage is vS = 5 V and R1 = R2 = 1 kΩ.
2. Repeat when resistor R3 is in the circuit and its value is R3 = 1 k Ω.
3. Repeat when resistor R3 is in the circuit and its value is R3 = 0.1 k Ω.

Solution:
1. For series circuits, we can apply Voltage divider rule:

𝑅𝑅𝑛𝑛 𝑅𝑅2 1
𝑣𝑣𝑛𝑛 = 𝑣𝑣 𝑣𝑣𝑛𝑛 = 𝑣𝑣2 = 𝑣𝑣 = 5 = 2.50 𝑉𝑉
𝑅𝑅1 + 𝑅𝑅2 + ⋯ + 𝑅𝑅𝑛𝑛 𝑠𝑠 𝑅𝑅1 + 𝑅𝑅2 𝑠𝑠 1 + 1

2. For parallel circuit, when R3 = 1 kΩ:

1 1 𝑅𝑅2 ∥ 𝑅𝑅3 0.5

𝑅𝑅2 ∥ 𝑅𝑅3 = = = 0.5
1 1 1 1 𝑣𝑣𝑛𝑛 = 𝑣𝑣2 = 𝑣𝑣𝑠𝑠 = 5 = 1.67 𝑉𝑉
+
𝑅𝑅2 𝑅𝑅3 1
+
1 𝑅𝑅1 + 𝑅𝑅2 ∥ 𝑅𝑅3 1 + 0.5

3. For parallel circuit, when R3 = 0.1 kΩ:

𝑅𝑅2 ∥ 𝑅𝑅3 =
1 1 𝑅𝑅2 ∥ 𝑅𝑅3 0.091
1 1
=
1 1
= 0.091 𝑣𝑣𝑛𝑛 = 𝑣𝑣2 = 𝑣𝑣𝑠𝑠 = 5 = 0.4167 𝑉𝑉
+
𝑅𝑅2 𝑅𝑅3 1
+
0.1
𝑅𝑅1 + 𝑅𝑅2 ∥ 𝑅𝑅3 1 + 0.091

Lecture No.4 6
 Norton Equivalent Circuits Lecture .5
It is possible to view even a very complicated circuit in terms of much simpler equivalent source and load
circuits, and that the transformations leading to equivalent circuits are easily managed, with a little practice.
The Norton Theorem
When viewed from the load, any network composed of ideal voltage and current sources, and of linear resistors,
may be represented by an equivalent circuit consisting of an ideal current source iN in parallel with an equivalent
resistance RN .

Determination of Norton Equivalent Resistance

1. Remove the load.
2. Zero all independent voltage and current sources.
3. Compute the total resistance between load terminals, with the load removed. This resistance is equivalent to
that which would be encountered by a current source connected to the circuit in place of the load.

Lecture No.5 1
Example: Find the Norton equivalent resistance RT seen by the load RL in the circuit of Figure
below.
R1 = 20 Ω; R2 = 20 Ω; I = 5A;
R3 = 10 Ω; R4 = 20 Ω; R5 = 10 Ω.

Solution:
we first set the current source equal to zero, by replacing it with an open circuit.
Looking into terminal a-b, we recognize that, starting from the left (away from the load) and
moving to the right (toward the load), the equivalent resistance is given by the expression

1
𝑅𝑅1 ∥ 𝑅𝑅2 =
1/𝑅𝑅1 + 1/𝑅𝑅2

Lecture No.5 2
Example: Compute the Norton equivalent resistance RT seen by the load in the circuit of Figure ,
V = 5V; R1 = 2 Ω; R2 = 2 Ω; R3 = 1 Ω;
I = 1A, R4 = 2 Ω.

Solution:
We first set the current source equal to zero, by replacing it with an open circuit,
then set the voltage source equal to zero by replacing it with a short circuit.
Looking into terminal a-b, we recognize that, starting from the left (away from the load) and
moving to the right (toward the load), the equivalent resistance is given by the expression

1
𝑅𝑅1 ∥ 𝑅𝑅2 =
1/𝑅𝑅1 + 1/𝑅𝑅2
Lecture No.5 3
Computing the Norton Current
The Norton equivalent current is equal to the short-circuit current that would flow if the load were replaced by
a short circuit.

Computation of Norton current

Illustration of Norton equivalent circuit

COMPUTING THE NORTON CURRENT

1. Replace the load with a short circuit.
2. Define the short-circuit current iSC to be the Norton equivalent current.
3. Apply any preferred method (e.g., node analysis) to solve for iSC.
4. The Norton current is iN = iSC.

Lecture No.5 4
Example: Determine the Norton current and the Norton equivalent for the circuit of
Figure ,
Equivalent resistance RN ; Norton current iN = iSC.
V = 6 V; I = 2A; R1 = 6 Ω ; R2 = 3 Ω; R3 = 2 Ω .

Solution:
We first compute the Norton equivalent resistance. We zero the two sources by shorting the voltage source
and opening the current source. The resulting circuit is shown in Figure.
RN = R1 || R2 + R3 = 6 || 3 + 2 = 4 Ω .
RN = 4 Ω

Next we compute the Norton current.

we replace the load with a short circuit and label the
short-circuit current iSC.
The circuit is shown in Figure ready for node voltage
analysis.
Note that we have identified two node voltages
v1 and v2,
and that the voltage source requires that v2 − v1 = V.
The unknown current flowing through the voltage
source is labeled i.
Lecture No.5 5
Applying KCL at node 1, we obtain the following equation:

I i

i1

(assuming currents entering a node as being positive)

Lecture No.5 6
Applying KCL at node 2, we obtain the following equation:

i3

i2

(assuming currents entering a node as being positive)

Lecture No.5 7
Lecture No.5 8
Knowing that:
V = 6 V; I = 2A; R1 = 6 Ω ; R2 = 3 Ω; R3 = 2 Ω .

i+0.1667 v2 = 3 …..(1)
-i+0.8333 v2 = 0 …..(2)
∴ i = 0.8333 v2 …..(3)
Sub in (1):
0.8333 v2 + 0.1667 v2 = 3
∴ v2 = 3
Sub in (3):
i = 0.8333 (3) = 2.5 Norton equivalent circuit
Lecture No.5 9
MAXIMUM POWERTRANSFER Lecture .6
How much power can be transferred to the load from the source under the most ideal conditions?
Or, alternatively,
What is the value of the load resistance that will absorb maximum power from the source?

The answer to these questions is contained in the maximum power transfer theorem.
Consider that the power absorbed by the load PL is given by

and that the load current is given by the familiar expression

Combining the two expressions, we can compute the load power as

To find the value of RL that maximizes the expression for PL (assuming that VT and RT are fixed),
the simple maximization problem:

must be solved. Computing the derivative, we obtain the following expression:

which leads to the expression

It is easy to verify that the solution of this equation is

This analysis shows that to transfer maximum power to a load, given a fixed
equivalent source resistance, the load resistance must match the equivalent source
resistance.

When RL=RT , then by sub. in eq. PL= v2T /4 RT

When RL≠RT ,

Lecture No.6 2
Example: The circuit diagram of Figure shows dry cells of source voltage 6 V, and internal
resistance 2.5 Ω. If the load resistance RL is varied from 0 to 5 in 0.5 Ω steps, calculate the power
dissipated by the load in each case. Plot a graph of RL (horizontally) against power (vertically) and
determine the maximum power dissipated.
A graph of RL against P is shown in Figure. The maximum value of power is 3.60 W which
occurs when RL is 2.5 , i.e. maximum power occurs when RL = r, which is what the maximum
power transfer.
Example: A DC source has an open-circuit voltage of 30 V and an internal resistance
of 1.5 Ω . State the value of load resistance that gives maximum power dissipation and
determine the value of this power.

From the maximum power transfer theorem, for maximum power dissipation,

RL = r = 1.5 Ω

Lecture No.6 5
Capacitors and Inductors Lecture .7
There are two important passive linear circuit elements: the capacitor and the inductor.
Unlike resistors, which dissipate energy, capacitors and inductors do not dissipate but store energy,
which can be retrieved at a later time. For this reason, capacitors and inductors are called storage
elements. The application of resistive circuits is quite limited, while the application of capacitor &
inductor is wide.

A capacitor consists of two conducting plates separated by an insulator (or dielectric).

Capacitors are used extensively in electronics, communications, computers, and power systems.
In many practical applications, the plates may be
aluminum foil while the dielectric may be air,
ceramic, paper, or mica.

When a voltage source is connected to the

capacitor, the source deposits a positive charge
+q on one plate and a negative charge –q on the
other.
The capacitor is said to store the electric
charge. The amount of charge stored, A typical capacitor
represented by q, is directly proportional
to the applied voltage so that
A capacitor with applied voltage v.
q=Cv
where C, is known as the capacitance
of the capacitor. The unit of capacitance is the
farad (F). [[1 farad 1 coulomb/volt ]]
 This equation applies to only parallel-plate capacitors,

where A is the surface area of each plate,

d is the distance between the plates,
ɛ is the permittivity of the dielectric material between the plates.

three factors determine the value of the capacitance:

1. The surface area of the plates—the larger the area, the greater the capacitance.
2. The spacing between the plates—the smaller the spacing, the greater the capacitance.
3. The permittivity of the material—the higher the permittivity, the greater the capacitance.
Capacitors are commercially available in different values and types.
Typically, capacitors have values in the picofarad (pF) to microfarad (μF)range. They are
described by the dielectric material they are made of.

Circuit symbol for capacitor

(a) polyester capacitor, (b) ceramic capacitor, (c) electrolytic capacitor

To obtain the current-voltage relationship of the capacitor, we take the derivative of both sides of
Eq : q=Cv . Since : i = dq/dt then by differentiating both sides of eq. : i = C dv/dt
This is the current-voltage relationship for a capacitor.
The voltage-current relation of the capacitor
can be obtained by integrating both sides of Eq.
i = C dv/dt
We get:

where v(to) = q(t0) / C is the voltage across the capacitor at time to.

The energy stored in the capacitor is:

Where : q=Cv
Then:
Eq. represents the energy stored in the electric field that exists between the plates of the capacitor.
Example:

Example:
Inductors
An inductor is a passive element designed to store energy in its magnetic field. Inductors find
numerous applications in electronic and power systems. They are used in power supplies,
transformers, radios, TVs, radars, and electric motors.

If current is allowed to pass through an inductor, it is

found that the voltage across the inductor is directly
proportional to the time rate of change of the current.
where L is the constant of
proportionality called the
inductance of the
inductor. The unit of inductance is
the henry (H). Typical form of an inductor.

1 henry equals 1 volt-second per ampere.

The inductance of an inductor depends on its physical dimension and construction. Formulas for
calculating the inductance of inductors of different shapes are derived from electromagnetic theory
and can be found in standard electrical engineering handbooks. For example, for
the inductor, (solenoid) shown in Fig. above,

where N is the number of turns,

l is the length,
A is the cross-sectional area,
and μ is the permeability of the core.
Circuit symbols for
inductors:
(a) air-core,
(b) iron-core,

Various types of inductors:

(a) Solenoidal wound inductor,
(b) toroidal inductor,
(c) chip inductor.
The inductor is designed to store energy in its magnetic field. The energy stored can be obtained
from Eq. v=L di/dt The power delivered to the inductor is

The energy stored is

Example
Electric Motors Lecture .8
DC motors that convert DC electric energy to mechanical energy.
Advantages of DC motors: easy speed and torque regulation.
There are really two types of motors, AC and DC.
The basic principles are alike for both.
Magnetism is the basis for all electric motor operation.
It produces the force required to run the motor.
There are two types of magnets the permanent magnet and the electro magnet.
Electro magnets have the advantage over permanent magnet in that the magnetic field can
be made stronger. Also the polarity of the electro magnet can easily be reversed. The
construction of an electro magnet is simple.
When a current passes through a coil of wire, a magnetic field is produced.

This magnetic field can be made stronger by winding the coil of wire on an iron core.
One end of the electro magnet is a north pole and the other end is a south pole The poles can be
reversed by reversing the direction of the current in the coil of wire. Likewise, if you pass a coil of wire
through a magnetic field, a voltage will be induced into the coil And, if the coil is in a closed circuit, a
current will flow.
 DC Motor Operation
Basic mechanism of the DC Motor – The Electromagnetism.

Lines of flux define the

magnetic field and are in the
form of concentric circles
around the wire.

The magnetic lines around a

current carrying conductor
leave from the N-pole and re-
enter at the S-pole.

Fleming's left hand rule

"Left Hand Rule" states that if you
point the thumb of your left hand in
the direction of the current, your
fingers will point in the direction of
the magnetic field.
2
 Types of
Motors

LINEAR
INDUCTION

DC BRUSH

ELECTROSTATIC

UNIVERSAL
STEPPER 3
DC Motor

4
 DC Motor parts:

- STATOR : provides mechanical support for the machine, consists poles and yoke
- ROTOR / ARMATURE : the rotating part, shrouded by fixed poles on the stator
-COMMUTATOR : mechanical rectifier, which changes the AC voltage of the rotating conductors to DC voltage
- BRUSHES : conduct the current from the commutator to the external circuit
- WINDINGS
-In most electric motors, rotor and stator are made of highly magnetically permeable materials -
steel or iron.
DC machines have two sets of electrical windings:
1)field windings - on stator
2)amarture windings - on the rotor.

DC motor stator with poles visible Rotor of a dc motor

5
DC Motor Construction

6
DC Motor Operation

7
DC Motor Operation : Current

8
DC Motor Operation : Force

9
DC Motor Operation : Magnetic Field

10
 Motor Modeling

Current
I
V Voltage MOTOR τ ,ω
Torque
Speed
Q Heat Out

Power In = Power Out

VI = Q + τω
VI ≅ I R + τω
2

11
The torque (T) of the motor is determined by

T = Kt ϕF IA
Where:
T = Torque
Kt = Torque constant
ϕF = % of field flux
IA = Armature current

The rotational speed (N) of the motor is determined by the voltage applied to the
armature coil.

𝑬𝑬𝑻𝑻 − 𝑰𝑰𝑨𝑨 𝑹𝑹𝑨𝑨

𝑵𝑵 =
𝑲𝑲𝑽𝑽 ∅𝑭𝑭
Where:
N = speed (rev/min)
ET = terminal voltage
IA = armature current
RA = armature resistance
KV =voltage constant
ϕF = % of field flux

12
 Typical speed – torque characteristics of mechanical loads

P= T ω
𝒏𝒏
𝝎𝝎 = 𝟐𝟐𝟐𝟐
𝟔𝟔𝟔𝟔

Where:
P : mechanical power of the load.
T: torque
ω : angular speed (rad/sec)
n: speed (rev/min)

• Torque proportional to the square of speed: like fans, pumps.

• Torque inversely proportional to speed: like milling machines. This load usually
requires a large torque at starting and at low speeds. The power consumption is
independent of speed.
• Torque linearly dependent on speed: like motor driving a DC generator connected to
a fixed resistance load.

13
 Equivalent circuit of DC motor

Applying KVL we obtain the voltage equation as:

Multiplying equation by the current gives the power equation as:

Electrical input power (VI) = Mechanical output power (EI) + Copper loss (I2R)

Copper loss refers to heat generated by the current in the windings.

14
 Lecture No.9 Dr. Laith Abdullah Mohammed
DC Electric Motors / Back EMF

When a DC motor rotates, an EMF is induced in the armature conductors.

By Lenz’s law this induced EMF ( E ) opposes the supply voltage ( V ) and is called a
back EMF, and the supply voltage, V is given by:

V = E + IaRa OR E = V - IaRa
Example: A DC motor operates from a 240 V supply. The armature resistance is 0.2 Ω . Determine
the back EMF when the armature current is 50 A.
DC Electric Motors / Torque of DC machine

Example: Determine the torque developed by a 350 V DC motor having an armature resistance of
0.5Ω and running at 15 rev/s. The armature current is 60 A.

2
DC Electric Motors / Shunt – wound Motors
In the shunt wound motor the field winding is in parallel with the armature across the supply as
shown in Figure.

Example: A 240 V shunt motor takes a total current of 30 A. If the field winding resistance Rf = 150
Ω and the armature resistance Ra = 0.4 Ω determine (a) the current in the armature, and (b) the back
EMF

3
DC Electric Motors / The efficiency of DC Motor

4
Example: A 320 V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron,
friction and windage losses amount to 1.5 kW, the shunt field resistance is 40 Ω and the armature
resistance is 0.2 Ω , determine the overall efficiency of the motor.

5
 Lecture No.10
AC Motors: Three-phase Induction motors
The three-phase induction motor, also called an asynchronous motor, is the most commonly used type
of motor in industrial applications. In particular, the squirrel- cage design is the most widely used
electric motor in industrial applications.
The magnetic field rotates and this has the advantage that no external electrical connections to the rotor
need be made. Its name is derived from the fact that the current in the rotor is induced by the
magnetic field instead of being supplied through electrical connections to the supply.
The result is a motor which:
(i) is cheap,
(ii) is explosion proof, due to the absence of a commutator or slip-rings and brushes with their associated
sparking,
(iii) requires little or no skilled maintenance, and
(iv) has self-starting properties when switched to a supply with no additional expenditure on auxiliary
equipment.
The principal disadvantage of a three-phase induction motor is that its speed cannot be readily adjusted.
When a three-phase supply is connected to symmetrical three-phase windings, the currents flowing in
the windings produce a magnetic field. This magnetic field is constant in magnitude and rotates at
constant speed as shown below, and is called the synchronous speed.
the windings are represented by three
single-loop conductors, one for each phase,
marked RSRF, YSYF and BSBF,
the S and F signifying start and finish.

When the stator windings are connected to a three-phase

supply, the current flowing in each winding varies with time
and is as shown in Figure.
If the value of current in a winding is positive, the
assumption is made that it flows from start to finish of the
winding, i.e., if it is the red phase, current flows from RS to
RF.
When the value of current is negative, the assumption is
made that it flows from finish to start.

At time, say t1, shown in Figure, the current flowing in the

red phase is a maximum positive value.
At the same time, t1, the currents flowing in the yellow and
blue phases are both 0.5 times the maximum value and are
negative.

LECTURE NO.10 2
The current distribution in the stator windings is therefore as
shown in Figure, in which current flows (shown as  ) in RS
since it is positive, (shown as ʘ ) in YS and BS, since
these are negative.

A short time later at time t2, the current flowing in the red phase has
fallen to about 0.87 times its maximum value and is positive, the current
in the yellow phase is zero and the current in the blue phase is about
0.87 times its maximum value and is negative.

At time t3, the currents in the red and yellow phases are 0.5
of their maximum values and the current in the blue phase
is a maximum negative value.

LECTURE NO.10 3
Synchronous speed:
The rotating magnetic field produced by three phase windings could have been produced by rotating
a permanent magnet’s north and south pole at synchronous speed, (shown as N and S at the ends of
the flux in Figures previously. For this reason, it is called a 2-pole system and an induction motor
using three phase windings only is called a 2-pole induction motor.
In general, if f is the frequency of the currents in the stator windings
and the stator is wound to be equivalent to p pairs of poles,
the speed of revolution of the rotating magnetic field, i.e., the synchronous speed, ns is given by:

Example 1. A three-phase two-pole induction motor is connected to a 50 Hz supply. Determine the

synchronous speed of the motor in rev/min.

LECTURE NO.10 4
Construction of a three-phase induction motor
The stator of a three-phase induction motor is the stationary part corresponding to the yoke of a d.c.
machine.

It is wound to give a 2-pole, 4- pole, 6-pole, . . . . . . rotating magnetic field, depending on the rotor
speed required.

The rotor, corresponding to the armature of a d.c. machine, is built up of laminated iron, to reduce
eddy currents.
In the type most widely used, known as a squirrel-cage rotor, copper or aluminum bars are placed in
slots cut in the laminated iron, the ends of the bars being welded or brazed into a heavy conducting
ring, (see Figure).
A cross-sectional view of a three-phase induction motor is shown
in Figure.

Al or Cu Bar

LECTURE NO.10 5
 T = 9555 (P / N)
T : Rated Torque (N.m)
P : Power (kw)
N : Speed (rpm)
Note: A standard motor must be able to deliver the rated torque in continuous operation without exceeding its
temperature limit.

Load torques as a function of speed

Torque remains constant: P ∝ N
Example: elevators, machine tools with a constant cutting force,
conveyor belts, piston pumps and compressors at constant pressure

Torque increases in proportion to speed: P ∝ N2

Example: rolling and processing of paper.

Torque increases with the square of speed: P ∝ N3

Example: pump and fan motors for heating and ventilation motors

Torque decreases in inverse proportion to speed: P ≅ constant

Example: winding machines, facing turning.

6
 Lecture No.11
Step Motor

Typical Applications
• Automation and inspection
• Conveyor transfer
• Cutting metal, plastic, fabric, etc.
• Material handling
• Medical equipment
• Office peripheral equipment (Printer, Plotter, Fax)
• Packaging systems
• Pick-and-place applications
• Robotics
• Semiconductor manufacturing
A stepper motor is an electro mechanical device, which converts electrical pulses into
discrete mechanical movements.
The shaft or spindle of a stepper motor rotates in discrete step increments when electrical command
pulses are applied to it in the proper sequence.
The sequence of the applied pulses is directly related to the direction of motor shafts rotation. The
speed of the motor shafts rotation is directly related to the frequency of the input pulses and the length
of rotation of input pulses applied.
Stepping motors can be controlled directly by computers or microcontrollers.

Advantages:
1. The rotation angle of the motor is proportional to the input pulse.
2. The motor has full torque at stand still (if the winding are energized)
3. Precise positioning: accuracy of 3 –5% of a step and this error is non cumulative from one step to the next.
4. Excellent response to starting stopping reversing.
5. Very reliable since there are no contact brushes in the motor. Therefore the life to the motor is simply dependant on
the life of the bearing.
6. The motors response to digital input pulses provides open-loop control, making the motor simpler and less costly to
control.
7. It is possible to achieve very low speed synchronous rotation with a load that is directly coupled to the shaft.
8. A wide range of rotational speed is proportional to the frequency of the input pulses.

Disadvantages
1. Resonance can occur if not properly controlled.
2. Not easy to operate at extremely high speeds.

Lecture No.11 2
There are three basic stepper motor types:
• Variable-reluctance
• Permanent-magnet
• Hybrid
Variable-reluctance (VR)
It is the easiest type. It consists of a soft iron multi-toothed rotor and
wound stator. When the stator windings are energized with DC current the
poles become magnetized. Rotation occurs when the rotor teeth are
attracted to energized stator poles.

Classification features of Step Motor:

1. Step Angle.
2. Frame Size: related to Body diameter of motor. (for example: size 23
step motor means motor have body diameter 2.3 inch).
3. Power: range (1<P<20 watt)
Cross-section of a
variable reluctance (VR) motor
Step angles are mostly in the range 1.88–90º, with torques ranging from 1µNm (in a
tiny wristwatch motor of 3 mm diameter) up to perhaps 40 Nm in a motor of 15 cm
diameter suitable for a machine tool application where speeds of 500 rev/min .

Lecture No.11 3
Animation of a simplified stepper motor (unipolar)
Frame 1: The top electromagnet (1) is turned on, attracting the nearest teeth of the gear-
shaped iron rotor. With the teeth aligned to electromagnet 1, they will be slightly offset from
right electromagnet (2).
Frame 2: The top electromagnet (1) is turned off, and the right electromagnet (2) is energized,
pulling the teeth into alignment with it. This results in a rotation of 3.6° in this example.
Frame 3: The bottom electromagnet (3) is energized; another 3.6° rotation occurs.
Frame 4: The left electromagnet (4) is energized, rotating again by 3.6°. When the top
electromagnet (1) is again enabled, the rotor will have rotated by one tooth position; since
there are 25 teeth, it will take 100 steps to make a full rotation in this example.
Lecture No.11 4
 Torque Generation
The torque produced by a stepper motor depends on several factors.
•  step rate
•  drive current in the windings
•  drive design or type

The torque is developed when the magnetic fluxes of the rotor and stator are displaced from each other.
The stator is made up of high permeability magnetic material.
The flux concentrate at the stator poles. The torque output produced by the motor is
proportional to the intensity of the magnetic flux generated when the winding is energized

Torque ∝ Magnetic field Intensity.

The basic relationship which defines the intensity of the magnetic flux is defined by:
H=(N * i) / L
Where: N= Number of windings.
i= Current.
H= Magnetic field Intensity.
L= Magnetic Flux path Length.

Lecture No.11 5
Phase , Poles, Stepping Angle:
Usually stepper motors have two phases.
A bipolar motor with two phases has
one winding/phase
A unipolar motor has one winding,
with a center tap per phase.

A pole can be defined as one of the

regions in a magnetized body where the magnetic flux density is concentrated.
Both the rotor and the stator of a step motor have poles.
In reality several more poles are added to both the rotor and stator structure in order to
increase the number of steps per revolution of the motor, (provide a smaller basic stepping angle).

Rotor teeth = No of rotor poles.

Nt= Total No. of poles = rotor teeth * stator phases
∴ Step angle = 360º / Nt

Example: The VR motor has four rotor teeth, three stator phase windings and the step angle is
therefore 30º.

Lecture No.11 6
 Electric Drive System Lecture 12
A modern electric drive system has five main functional blocks:
1. Mechanical Load: determined by nature of the industrial operation.
2. Motor.
3. Converter: interface the motor with power source & provides the motor with adjustable
voltage, current, and /or frequency.
4. Power source: provides the energy to the drive system.
5. Controller: supervise the operation of the entire system to enhance overall system
performance and stability.

The designer can select the Electric Motor, Convertor, Controller.

An electric drive is the

electromechanical system that converts
electrical energy to mechanical motion

Functional diagram of electric drive

Lecture No.12 1
It includes a motor M (or several ones), a mechanical transmission (gear, gearbox), an optional power converter, and a
control system (controller). The power converter transforms electrical energy W0 of the grid (mains) to motor supply
energy W1 in response to the set-point speed or path command. The motor is an electromechanical converter, which
initially converts W1 to electromagnetic energy W12 of the air gap between the stator and the rotor and then turns W12 to
mechanical work W on the motor shaft. The gear transforms mechanical energy to the load work WL in accordance with
the requirements of the driven machine (actuator). The controller (regulator) compares the set-point y* with outputs y and
disturbances χ, and generates the references δ on its inputs. The part of electric drive, which involves the mechanical
transmission and the motor rotor, is called a mechanical system.

Applications:
Nowadays electric drives can be found nearly everywhere, in heating, ventilation and air conditioning,
compressors, washing machines, elevators, cranes, water pumping stations and wastewater processing plants, conveyors

Electric drives use approximately 70 % of generated electrical energy.

It is more than 100000 billions kilowatt-hours per year.
Electric drive systems make up about one-third of overall automation equipment.

The leading companies in the world market of electric drive engineering are now as follows:
• American General Electric,
• LabVolt,
• Canadian Allen Bradley;
• Siemens, Bosh,
• Danish Danfoss;
• ABB Brown Bowery,
• Japanese Fanuc,
• Mitsubishi Electric,
• Hitachi;
• SwissLecture No.12 Automation,
Rockwell 2
Power sources:
Two major types of Power sources are used in industrial applications:
1. Alternating Current (AC): can either be single or multi phase system. Single phase systems are common
where the demand for electric power is limited. Multiphase are used in high power consumption
applications.
2. Direct Current (DC).

In US 3phase 60 Hz power sources is common.

In Europe, Middle East, Africa, Asia, frequency is 50 Hz.

Controllers:
Monitor system variables, compare them with some desired values, and then readjust the converter
output until the system achieve a desired performance. Controllers used in speed or position control.

Converters:
Transform the waveform of power sources to that required by an electric motor.
Most converters provide adjustable voltage, current, frequency to control speed, torque, power of motor.

Lecture No.12 3
 Four Types of Convertors

← Suitable for Induction Motors.

← Suitable for DC Motors.

Called “Chopper”
Convert constant DC waveform to
variable magnitude DC waveform.

← Suitable for DC drives.

← Suitable for AC Motors.

Lecture No.12 4
Mechanical Loads:
Mechanical loads exhibit wide variations of speed – torque features. Load torques are generally speed
dependent and can be represented by formula:

The mechanical power of the load is given by:

P= T w , w= 2π * (n/60)
where: w: angular speed in rad/sec
n: speed in rev/min.

Typical speed – Torque chart

of mechanical loads.

At k=1: T∝ n , P ∝ n2
At k=0, C=1: T independent of n , P ∝ n
At k=2: T ∝ n2 , P ∝ n3
At k=-1: T ∝ 1/n , P independent of n

Lecture No.12 5
Typical speed – Power chart of mechanical loads.
CHOPPER-FED D.C. MOTOR DRIVES Lecture 13
• Chopper is a static device.
• A variable dc voltage is obtained from a constant dc voltage source.
• Also known as DC-to-DC converter.
• Widely used for motor variable speed control.
v0
V
Chopper
i0 Vdc
+
t
tON tOFF
V R V0 i0

− V/R
The transistor in the circuit acts as a switch. Idc
When transistor is ON, supply voltage appears
across the load
t
When transistor is OFF, the voltage across the load
T
will be zero.
Typical waveforms of armature voltage and current

LECTURE NO.13 1
Vdc = Average value of output or load voltage.
I dc = Average value of output or load current.
tON = Time interval for which SCR conducts.
tOFF = Time interval for which SCR is OFF.
T =tON + tOFF =Period of switching or chopping period.
1
f= = Freq. of chopper switching or chopping freq.
T
Average Output Voltage
 tON 
Vdc = V  
t +
 ON OFF 
t
Effective input resistance of chopper
 tON 
=Vdc V=   V .d V
 T  Ri =
I dc
t 
= d= duty cycle
but  ON R
Ri =
 t  d
The output voltage can be varied by
varying the duty cycle. 2
 The speed of the motor is determined by the average armature voltage, (Vdc), which in turn depends on
the proportion of the total cycle time (T) for which the transistor is ‘on’. If the on and off times are
defined as:
Ton = kT
Toff = (1 - k) T,
where 0 < k < 1, then the average voltage is simply given by Vdc = kV

[[ Speed control is effected via the on time ratio, k ]]

[[ Average current (Idc) produces the full rated torque of the motor]].

Methods Of Control
V0
The output DC voltage can be varied by using Pulse Width Modulation
control (PWM). V
tON is varied keeping chopping frequency ‘f’ and chopping period ‘T’
tON tOFF
constant.
Output voltage is varied by varying the ON time tON
t
T
V0

t
tON tOFF

LECTURE NO.13 3
Example 1: A Chopper circuit is operating on a DC drive system at a frequency of 2 kHz on a 460 V supply. If the load
voltage is 350 volts, calculate the conduction period of the transistor in each cycle.

V = 460 V, Vdc = 350 V, f = 2 kHz

1
Chopping period T=
f
1
=T = −3
0.5 m sec
2 ×10
t 
Output voltage Vdc =  ON  V
 T 
Conduction
Conduction periodperiod of thyristor
of transistor:

T × Vdc
tON =
V
0.5 ×10−3 × 350
tON =
460
tON = 0.38 msec
LECTURE NO.13 4
Example 2: A dc chopper has a resistive load of 20Ω and input voltage VS = 220V. When chopper is ON, its voltage
drop is 1.5 volts and chopping frequency is 10 kHz. If the duty cycle is 80%,
Determine the average output voltage and the chopper on time.

VS =220V , R =20Ω, f =10 kHz

tON
=
d = 0.80
T
Vch = Voltage drop across chopper = 1.5 volts
Average output voltage
t 
=Vdc  ON  (VS − Vch )
 T 
Vdc 0.80 ( 220 − 1.5
= = ) 174.8 Volts
Chopper ON time, tON = dT
1
Chopping period, T=
f
1
T= 3= 100 μsecs
0.1×10−3 secs =
10 ×10
Chopper ON time,
tON = dT
tON = 0.80 × 0.1× 10−3
tON = 0.08 × 10−3 = 80 μsecs 5
Example 3: In a DC chopper, the average load current is 30 Amps, chopping frequency is 250 Hz, supply
voltage is 110 volts. Calculate the ON and OFF periods of the chopper if the load resistance is 2 ohms.

I dc = 30 Amps, f = 250 Hz , V = 110 V , R = 2Ω

1 1
Chopping period, T = = = 4 ×10−3 =
4 msecs
f 250
Vdc
= I dc = & Vdc dV
R
dV
∴ I dc =
R
I dc R 30 × 2
= d = = 0.545
V 110
Chopper ON period,
tON = dT= 0.545 × 4 ×10−3= 2.18 msecs
Chopper OFF period,
tOFF= T − tON
4 ×10−3 − 2.18 ×10−3
tOFF =
tOFF = 1.82 ×10−3 = 1.82 msec 6
 Dynamic of Electric Drive Systems Lecture 14

In modern drives like robotics, controlling motor final speed is not the only goal.
Moving robot arm according to pre selected trajectories needs to control the traveling time of the motors during starting,
breaking and change of speed.
Traveling time: the time required to change the motor speed from one steady state operation point to another.
For example: for induction motor, the developing inertia torque is shown below.

Td: motor torque.

Tl: load torque.
Ti: inertia torque
w: motor speed

Ti=Td-Tl= J * (dw/dt)

When motor speed changes from w1 to w2 ,

Traveling time = t

From this equation: traveling time between two speed can be reduced when:
• The inertia torque increases (by increasing the induction motor voltage, leads to increases Ti
which is equal to Td-Tl))
• The moment of inertia (J) decreases (using gears or belt systems)

LECTURE NO.14 1
Moment of Inertia (J)

LECTURE NO.14 2
 LOAD

LECTURE NO.14 3
A system with linear and rotating motion

Connecting Rod

Piston
Motor wheel

LECTURE NO.14
 Belt Drive System

Gear Drive System

Gear system consist of a motor, mechanical load and a gear.

Gear ratio = (n1/n2) = (d2/d1)
n= shaft speed,
d= gear diameter.

LECTURE NO.14 5
LECTURE NO.14 6
LECTURE NO.14 7
Example:

J= Jm + Jl
Jm: Motor moment of inertia
Jl: Load moment of inertia

Ra: armature resistance (Ω)

Kφ = field constant of the
motor
tst : starting time of the
motor.

LECTURE NO.14 8
 Servo Motors Lecture 15
The servo type intended for use in applications which require rapid acceleration and deceleration.

The design of the motor will reflect this by catering for intermittent currents (and hence torques) of many times the
continuously rated value.

Because most servo motors are small, their armature resistances are relatively high:
In order to maximise acceleration, the rotor inertia must be minimised, and one obvious way to achieve this is to
construct a motor in which only the electric circuit (conductors) on the rotor move, the magnetic part
(either iron or permanent magnet) remaining stationary. This principle is adopted in ‘ironless rotor’ and ‘printed
armature’ motors.

In the ironless rotor , the armature conductors are formed as a thin-walled cylinder consisting essentially of
nothing more than varnished wires wound in skewed form together with the disc-type commutator . Inside the
armature sits a 2-pole (upper N, lower S) permanent magnet, which provides the radial flux, and outside it is a steel
cylindrical shell which completes the magnetic circuit.
The absence of slots to support the armature winding results in a relatively fragile structure, which is therefore limited
to diameters of not much over 1 cm. Because of their small size they are often known as micro motors, and are very
widely used in cameras, video systems, card readers etc.

High
Torque Ironless rotor DC motor
Motor

High Speed Spindle 1

The printed armature type is more robust, and is made in sizes up to a few kilowatts.
They are generally made in disc, with the direction of flux axial and the armature current radial.
The armature conductors resemble spokes on a wheel; the conductors themselves being formed on a lightweight disc.
Since there are usually at least 100 armature conductors, the torque remains almost constant as the rotor turns, which
allows them to produce very smooth rotation at low speed.
Inertia and armature inductance are low, giving a good dynamic response, and the short shape makes them suitable for
applications such as machine tools and disc drives.

Servo motors used in

feed drives: control
the position between
working piece and
working tool
main drives: provide
processing power

LECTURE NO.15 Typical Machine2 Tool

Open-Loop Control
It is the simplest form of drive control. The drive operates the motor,
given the speed reference that is delivered from the hand speed pot.
Open-loop control is more typical in AC drives.
DC drives are almost never operated open loop because of the higher
regulation characteristics.

Closed-Loop Control
this type of control method requires the use of some peripheral device that closes the loop back to the drive. The device
that is commonly used is the tachometer, or tach generator.
This control type yields a great improvement in motor-speed regulation.
A speed reference is given to the AC or DC drive. The actual speed, determined by the tach, is fed back to the
drive. Both the speed reference and feedback are combined into the summing circuit. The summing circuit produces an
error signal. That error signal is now the speed reference that is sent to the power and logic control, which translates the
signal into faster or slower motor speed.

LECTURE NO.15 3
If this happened to be an AC drive configured in torque-control mode, the feedback would be a current feedback
instead of speed. If this was a position control, such as a servo controller, the feedback would be a position and use an
encoder instead of a tach. The idea of this entire scheme is to have zero error on the output of the summing circuit.
If error occurs, it must be corrected. How fast it is corrected is a function of the logic control in the drive.

Tachometers
A tachometer generator or tach is an electromechanical device that translates the rotational speed of a shaft to an
electrical signal. An analog tach generator is a small generator that produces an output voltage whose magnitude is
linearly proportional to shaft speed.
An AC tach produces an AC voltage and a DC tach produces a DC voltage.
DC tachs produce a negative voltage to indicate reverse rotation as shown in Figure.
DC tachs are the most popular type of analog tach, since AC tachs do not indicate direction and are generally less
accurate than DC tachs. The output of a DC tach is typically specified in volts per 1000 rpm.
Digital tachs, produce a series of pulses. The frequency of those pulses is proportional to shaft speed. The output of a
digital tach is typically specified in pulses per revolution (PPR).

Digital tach output

DC tach output voltage vs. speed

4
 DC Tachogenerator Wiring diagram

Tachometer Tachogenerator

LECTURE NO.15 5
Resolvers
A resolver is a “position” transducer, with characteristics that resemble a small motor. This type of feedback device
would be used mainly with servo motor applications where precise feedback of rotor position is critical
to system accuracy.
For example, in cut-to-length applications, linear position of a sliding table or cutting arm would be a function of the
position of the rotor. Figure shows a diagram of a resolver.
A resolver is very similar to an AC induction motor. A resolver contains a single winding rotor that rotates inside fixed
coils of wire, called stators. A reference voltage is typically applied to the rotor winding. This rotating
winding has a magnetic field that induces a voltage in the stator windings, which produce an analog output. This analog
voltage output is proportional to shaft (rotor) rotation.
Basically, the resolver is a rotating transformer. The rotating primary (rotor) induces a voltage in the fixed winding
secondaries (stator windings).
The analog output is what is fed back to the drive as an actual speed signal, or what would be considered a position
signal.
Because there are no electronics involved with resolvers, they are better suited for dirty environments than encoders.
Also, the resolver is a device that is an absolute measuring instrument. It can retain its exact location during a power
outage. Typically the resolver can transmit information over distances up to 300 m with little effect from electrical
noise. The resolution of some resolvers can be rated as high as 16,384 counts (14 bit or 214).

LECTURE NO.15 6
The magnitude of the energy through the resolver windings
varies sinusoidally as the shaft rotates. A resolver control
transmitter has one primary winding, the Reference
Winding, and two secondary windings, the SIN and COS
Windings. (See figure). The Reference Winding is located in
the rotor of the resolver, the SIN and COS Windings in the
stator. The SIN and COS Windings are mechanically displaced
90 degrees from each other. In a brushless resolver, energy
is supplied to the Reference Winding (rotor) through a
rotary transformer. This eliminates brushes and slip rings in
the resolver and the reliability problems associated with
them.

In general, in a control transmitter, the Reference Winding is excited by an AC

voltage called the Reference Voltage (Vr). The induced voltages in the SIN and COS
Windings are equal to the value of the Reference Voltage multiplied by the SIN or
COS of the angle of the input shaft from a fixed zero point. Thus, the resolver
provides two voltages whose ratio represents the absolute position of the input
shaft. (SIN θ / COS θ = TAN θ, where θ = shaft angle.)
An additional advantage of this SIN / COS ratio is that the shaft angle is absolute.
Even if the shaft is rotated with power removed, the resolver will report its new
position value when power is restored.

LECTURE NO.15 7