University of Kufa Fourth Year, Introduction

Faculty of Engineering Electrical Drive and Control

Electrical Engineering Department Ali Qasim AL-Mousawi

REVIEW OF ELECTRIC DRIVES

Introduction to Electrical Drive: -
The first electric drive was invented in 1838 by B. S. Iakobi in Russia. He tested a
DC motor which is supplied from a battery to push a boat. Although, the application
of electric drive in industrial can happen after so many years like in 1870. At present,
this can be observed almost everywhere. We know that the speed of an electrical
machine (motor or generator) can be controlled by the source current’s frequency
as well as the applied voltage. Although, the revolution speed of a machine can also
be controlled accurately by applying the electric drive concept.

1.1 What is an Electric Drive?

An Electric Drive can be defined as, a system which is used to control the movement
of an electrical machine. This drive employs a prime mover such as a petrol engine,
otherwise diesel, steam turbines otherwise gas, electrical & hydraulic motors like a
main source of energy. These prime movers will supply the mechanical energy
toward the drive for controlling motion. At present, the controlling can be done
simply using the software. Thus, the controlling turns into more accurate & this drive
concept also offers the ease of utilizing. Electrical drives have many advantages
over other mechanical drives. Electrical drives are more popular for its simple
constructions, reliability, cleanliness, and smooth easy speed control.

The types of electrical drives are two such as a standard inverter (used to control the
torque & speed) as well as a servo drive (used to control the torque as well as speed,
and also components of the positioning machine utilized within applications that need
difficult motion).
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University of Kufa Fourth Year, Introduction
Faculty of Engineering Electrical Drive and Control
Electrical Engineering Department Ali Qasim AL-Mousawi

Component of Electric Drive

The electric drive can be built with source, power modulator, motor, load, sensing unit, control unit,
an input command.

1- Power Source
The power source offers the necessary energy for the system. The converter and the motor
interfaces by the power source to provide changeable voltage, frequency and current to the motor.
Very low power drives are generally fed from single phase sources. Rest of the drives is powered
from a 3-phase source. Low and medium power motors are fed from a 400v supply. For higher
ratings, motors may be rated at 3.3KV, 6.6KV and 11 KV. Some drives are powered from battery.

2- Power Modulator
can be used to control the output power of the supply. The power controlling of the motor can
be done when electrical motor sends out the speed-torque feature which is necessary with the load.
During the temporary operations, the extreme current will be drawn from the power source and may
excess it otherwise can cause a voltage drop. Therefore, the power modulator limits the motor
current. The power modulator can change the energy based on the motor requirement.
It can classify as: Controlled rectifier (AC/DC), Inverters (DC/AC), AC voltage controllers
(AC/AC), DC choppers (DC/DC), and Cyclo-converters (Frequency conversion).

3- Control Unit
used to control the power modulator, and this modulator can operate at power levels as well as
small voltage. And it also works the power modulator as preferred. This unit produces the rules for
the safety of the motor as well as power modulator. The input control signal regulates the drive’s
working point from input toward the control unit.

4- Sensing Unit
The sensing unit in the block diagram is used to sense the particular drive factor. This unit is
mainly used for the operation of closed loop otherwise protection. There are many types of sensors:-
Speed Sensing (From Motor)
Torque Sensing
Position Sensing
Current and Voltage Sensing from Lines or from motor terminals.
Torque sensing (from load). Temperature Sensing (from load).

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University of Kufa Fourth Year, Introduction
Faculty of Engineering Electrical Drive and Control
Electrical Engineering Department Ali Qasim AL-Mousawi

5- Motor
Convert energy from electrical to mechanical. In braking mode, the flow of power is reversed.
The electric motor can be chosen by believing various features such as price, power rating &
performance necessary by the load throughout the stable state as well as active operations.
The motor speed is determined mainly by the applied frequency. The motor slows down a little
as the load increases and the slip increases. If the load is too great the motor will exceed the
maximum torque and stall or “pull out‟. Most motors and inverters will operate at 150% load.
The losses in the machines contribute to the temperature increase in the machine. Allowable
power losses are higher for materials which can withstand higher temperature which translates to
higher costs. Three main cause of power losses are:
 Conductor losses: 𝑖 𝑅. Exist in the windings, cables, brushes, slip rings, commutator, and etc.
 Core losses: Mainly due to eddy current and hysteresis losses
 Friction and windage losses: Mainly due to ball bearings, brushes, ventilation losses

6- Load
The mechanical load can be decided by the environment of the industrial process & the power
source can be decided by an available source at the place. However, we can choose the
other electric components namely electric motor, controller, & converter. The load requirements
are in either of Speed control or Torque control.
Depending upon the load requirements the motor has to be chosen. For example, in traction
system the load needs high starting torque (initial i.e., high current value is needed at the start. A
series motor provides a high starting torque. Hence series motor should be chosen for traction
system). There are three types of industrial loads under which electric motors are required to work.
1) Continuous load: Load is continuous in nature. Ex- Pumps or fans require a constant power
input to keep them operating.
2) Intermittent load: This type classified in to two types: Motor loaded for short time and then
shunt off for sufficiently longer duration temperature is brought to the room temperature.
Ex: kitchen mixite.
3) Variable or fluctuating load.

Advantages of Electrical Drives:

 It has very large range of speed and power.
 It is independent of the environmental condition.
 It is readily available by means of electric power.
 They are pollution free as there are no flue gases in that.
 It operates on all the quadrants of speed torque plane.
 It can be started easily.
 It has high efficiency due to fewer losses.
 They have a longer life span than other drives systems.
 No need of any fuel storage and transportation.
 It has High efficiency.
 They require less space.
 It is reliable, economical source of power, and can be remotely controlled.

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University of Kufa Fourth Year, Introduction
Faculty of Engineering Electrical Drive and Control
Electrical Engineering Department Ali Qasim AL-Mousawi

Classification of Electrical Drives

The drives categorized based on the different parameters as shown below: -
 Based on supply namely AC drives & DC drives.
 Based on running speed namely Constant speed drives & changeable speed drives.
 Based on a number of motors namely Single motor drives & multi-motor drives.
 Based on control parameter namely stable torque drives & stable power drives.

In electrical drives choice, AC drive is preferred because of it cost less, AC power can be
transmitted with low line losses, and increase or decrease voltage without more losses of power.

DC Drives AC Drives
1 Power circuit and control circuit is simple Power and Control circuit is complex
2 Frequent Maintenance. Less Maintenance.
3 Commulator makes bulky, costly and Problems are not there, particularly squirrel
heavy. cage motor.
4 Speed and design rating are limited due to Ratings have no upper limits.
commutation.
5 This is used in certain location Used in all location
6 Fast response and wide speed range In solid state control speed range is wide and
smooth achieved by conventional and conventional method it is stepped and limited.
solid-state control
7 Poor PF, harmonic distortion of the For Regenerative drives the line pf is poor, for
current. non-regenerative drives the line PF is better.
8 Power / weight ratio is small. Power / weight ratio is large.

Classification of Electric Drives according to number of machines

The choice of the electric drives, there are three classification namely
 grope drive
 individual drive
 multimotor drive

Individual Drive
In individual drive, each individual machine is driven by a separate motor. This motor also
imparts motion to various other parts of the machine. Examples of such machines are single spindle
drilling machines (Universal motor is used) and lathes. In a lathe, the motor rotates the spindle,
moves the feed and also with the help of gears, transmits motion to lubricating and cooling pumps.
A three-phase squirrel cage induction motor is used as the drive. In many such applications the
electric motor forms an integral part of the machine.

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University of Kufa Fourth Year, Introduction
Faculty of Engineering Electrical Drive and Control
Electrical Engineering Department Ali Qasim AL-Mousawi

Advantages:
1. Machines can be located at convenient places.
2. Continuity in the production of the processing industry is ensured to a high level of reliability.
3. when is a fault in one motor, the effect on the production of the industry will not be appreciable.
Disadvantages: Initial cost is very high.

Group Drive
One motor is used as a drive for two or more than machines. The motor is connected to a long shaft.
All the other machines are connected to this shaft through belt and pulleys.

Advantages:
1. economical because the rating of the motor used less than the aggregate of the individual motors
required to drive each equipment, because all of they may not be working simultaneously.
2. reduces the initial cost of installing a particular industry.
3. Less space is required in group drive as compared to individual drive.
4. Low maintenance cost.

Disadvantages: The use of this kind of drive is restricted due to the following reasons:
1. Flexibility of lay out is lost and the possibility of installation of additional machines in an
existing industry is limited.
2. In case of any fault to the main driving motor, all the other motors will be stopped immediately.
3. Level of noise produced at the site is high.
4. This kind of drive is untidy appearance, and it is also less safe to operate.
5. power loss is high because of the large amount of energy is wasted in belts and pulleys.
6. Power factor: Group drive has low power factor
7. Speed: Group drive does not provide constant speed.
8. Group drive is not suitable for driving heavy machines such as cranes, and hoists.

Multimotor Drive
In this type of drive, separate motors are provided for actuating different parts of the driven
mechanism. Ex: cranes, drives used in paper mills, rolling mills etc.,
In cranes, separate motors are used for hoisting, long travel motion and cross travel motion.

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 University of Kufa Fourth Year, Introduction
Faculty of Engineering Electrical Drive and Control
Electrical Engineering Department Ali Qasim AL-Mousawi

Selecting a Drive
Often drive selection is straightforward, as a motor is already installed and the speed range
requirement is not excessive. However, when a drive system is selected from first principles, careful
consideration may avoid problems in installation and operation, and may also save significant cost.
and some considerations must be taken into account such as: -
a) Check the Current rating of the inverter and the motor. Power rating is only a rough guide.
b) Check that you have selected the correct operating voltage. 230V three phase input
MICROMASTERs will operate with single or three phase inputs; MIDIMASTERs will operate
with three phase only. Single phase input units can be more cost effective in some cases, but
note that 230V units will be damaged if operated at 400V.
c) Check the speed range you require. Operation above normal supply frequency (50 or 60Hz) is
usually only possible at reduced power. Operation at low frequency and high torque can cause
the motor to overheat due to lack of cooling.
d) Synchronous motors require de-rating, typically by 2 -3 times. This is because the power factor,
and hence the current, can be very high at low frequency.
e) Check overload performance. The inverter will limit current to 150 or 200 % of full current very
quickly - a standard, fixed speed motor will tolerate these overloads.
f) Do you need to stop quickly? If so, consider using a braking resistor to absorb the energy.
g) Do you need to operate with cables longer than 50m, or screened or armored cables longer than
25m? If so, it may be necessary to de-rate, or fit a choke to compensate for the cable
capacitance.
Choice (or) Selection of Electrical Drives
Choice of an electric drive depends on a number of factors. Some of the important factors are.
1. Steady State Operating conditions requirements
Nature of speed torque characteristics, speed regulation, speed range, efficiency, duty
cycle, quadrants of operation, speed fluctuations if any, ratings etc
2. Transient operation requirements
Values of acceleration and deceleration, starting, braking and reversing performance.
3. Requirements related to the source
Types of source and its capacity, magnitude of voltage, voltage fluctuations, power
factor, harmonics and their effect on other loads, ability to accept regenerative power
4. Capital and running cost, maintenance needs life.
5. Space and weight restriction if any.
6. Environment and location.
7. Reliability.
Questions and Answers
1. Mention the types of braking
Ans: - 1) Regenerative braking
2) Dynamic braking
3) Plugging
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University of Kufa Fourth Year, Introduction
Faculty of Engineering Electrical Drive and Control
Electrical Engineering Department Ali Qasim AL-Mousawi

2. What is meant by regenerative braking?

Ans: - Regenerative braking occurs when the motor speed exceeds the synchronous speed. In this case
the IM runs as the induction m\c is converting the mechanical power into electrical power which is
delivered back to the electrical system. This method of braking is known as regenerative braking.

3. What is meant by dynamic braking?

Ans: - Dynamic braking of electric motors occurs when the energy stored in the rotating mass is
dissipated in an electrical resistance. This requires a motor to operate as a gen. to convert the stored
energy into electrical.

4. What is meant by plugging?

Ans: - It is one method of braking of IM. When phase sequence of supply of the motor running at the
speed is reversed by interchanging connections of any two phases of stator with respect to supply
terminals, operation shifts from motoring to plugging region.

5. Which braking is suitable for reversing the motor?

Ans: - Plugging is suitable for reversing the motor.

6. What is critical speed?

Ans: - It is the speed that separates continuous conduction from discontinuous conduction mode.

7. Define equivalent current method.

Ans: - The motor selected should have a current rating more than or equal to the current. It is also
necessary to check the overload of the motor. This method of determining the power rating of the motor
is known as equivalent current method.

8. Define four quadrant operation.

Ans: - The motor operates in two mode: motoring and braking. In motoring, it converts electrical energy
into mechanical energy which supports its motion. In braking, it works as a generator, converting
mathematical energy into electrical energy and thus opposes the motion. Motor can provide motoring
and braking operations for both forward and reverse directions.

9. What is meant by mechanical characteristics?

Ans: The curve is drawn between speed and torque. This characteristic is called mechanical
characteristics.

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 UNIT-I
DRIVE CHARACTERISTICS

Electrical Drives:
Motion control is required in large number of industrial and domestic applications like
transportation systems, rolling mills, paper machines, textile mills, machine tools, fans, pumps, robots,
washing machines etc.
Systems employed for motion control are called DRIVES, and may employ any of prime
movers such as diesel or petrol engines, gas or steam turbines, steam engines, hydraulic motors and
electric motors, for supplying mechanical energy for motion control. Drives employing electric motors
are known as ELECTRICAL DRIVES.
An ELECTRIC DRIVE can be defined as an electromechanical device for converting electrical
energy into mechanical energy to impart motion to different machines and mechanisms for various
kinds of process control.
Applications

Paper mills
Steel Mills

Cement Mills
Electric Traction

Textile mills
Petrochemical Industries

Sugar Mills
Electrical Vehicles

Multi quadrant Operation:

For consideration of multi quadrant operation of drives, it is useful to establish suitable
conventions about the signs of torque and speed. A motor operates in two modes – Motoring and
braking. In motoring, it converts electrical energy into mechanical energy, which supports its motion.
In braking it works as a generator converting mechanical energy into electrical energy and thus
opposes the motion. Motor can provide motoring and braking operations for both forward and reverse
directions. Figure shows the torque and speed co-ordinates for both forward and reverse motions.
Power developed by a motor is given by the product of speed and torque. For motoring operations
power developed is positive and for braking operations power developed is negative.
Speed
Forward Forward
Braking Motoring

II I

Torque
III IV
Reverse Reverse
Motoring Braking
 In quadrant I, developed power is positive, hence machine works as a motor supplying
mechanical energy. Operation in quadrant I is therefore called Forward Motoring. In quadrant II,
power developed is negative. Hence, machine works under braking opposing the motion. Therefore
operation in quadrant II is known as forward braking. Similarly operation in quadrant III and IV can
be identified as reverse motoring and reverse braking since speed in these quadrants is negative. For
better understanding of the above notations, let us consider operation of hoist in four quadrants as
shown in the figure. Direction of motor and load torques and direction of speed are marked by arrows.

T
Tl
T
ωm Tl
ωm

Motion
Motion

Counter
weight Empty Counter
Cage II I weight Loaded
Cage

T III IV T
Tl Tl
ωm
ωm
Motion
Motion

Empty
Cage Loaded
Cage

Counter
weight
Counter
weight Load Torque with Load Torque with loaded
empty cage cage
 A hoist consists of a rope wound on a drum coupled to the motor shaft one end of the rope is
tied to a cage which is used to transport man or material from one level to another level . Other end of
the rope has a counter weight. Weight of the counter weight is chosen to be higher than the weight of
empty cage but lower than of a fully loaded cage. Forward direction of motor speed will be one which
gives upward motion of the cage. Load torque line in quadrants I and IV represents speed-torque
characteristics of the loaded hoist. This torque is the difference of torques due to loaded hoist and
counter weight.
The load torque in quadrants II and III is the speed torque characteristics for an empty hoist.
This torque is the difference of torques due to counter weight and the empty hoist. Its sigh is negative
because the counter weight is always higher than that of an empty cage.
The quadrant I operation of a hoist requires movement of cage upward, which corresponds to
the positive motor speed which is in counter clockwise direction here. This motion will be obtained if
the motor products positive torque in CCW direction equal to the magnitude of load torque TL1. Since
developed power is positive, this is forward motoring operation. Quadrant IV is obtained when a
loaded cage is lowered. Since the weight of the loaded cage is higher than that of the counter weight .It
is able to overcome due to gravity itself.
In order to limit the cage within a safe value, motor must produce a positive torque T equal to
TL2 in anticlockwise direction. As both power and speed are negative, drive is operating in reverse
braking operation. Operation in quadrant II is obtained when an empty cage is moved up. Since a
counter weigh is heavier than an empty cage, its able to pull it up. In order to limit the speed within a
safe value, motor must produce a braking torque equal to TL2 in clockwise direction. Since speed is
positive and developed power is negative, it’s forward braking operation.
Operation in quadrant III is obtained when an empty cage is lowered. Since an empty cage has
a lesser weight than a counter weight, the motor should produce a torque in CW direction. Since speed
is negative and developed power is positive, this is reverse motoring operation.
Steady State Stability:
Equilibrium speed of motor-load system can be obtained when motor torque equals the load
torque. Electric drive system will operate in steady state at this speed, provided it is the speed of stable
state equilibrium. Concept of steady state stability has been developed to readily evaluate the stability
of an equilibrium point from the steady state speed torque curves of the motor and load system.
 In most of the electrical drives, the electrical time constant of the motor is negligible compared
with the mechanical time constant. During transient condition, electrical motor can be assumed to be
in electrical equilibrium implying that steady state speed torque curves are also applicable to the
transient state operation.
Now, consider the steady state equilibrium point A shown in figure below

ωm
T
TL

A
∆ωm

Tshift TA TM Torque

The equilibrium point will be termed as stable state when the operation will be restored to it
after a small departure from it due to disturbance in the motor or load. Due to disturbance a reduction
of ∆ωm in speed at new speed, electrical motor torque is greater than the load torque, consequently

motor will accelerate and operation will be restores to point A. similarly an increase in ∆ωm speed
caused by a disturbance will make load torque greater than the motor torque, resulting into
deceleration and restoring of operation to point A.

Now consider equilibrium point B which is obtained when the same motor drives another load
as shown in the figure. A decrease in speed causes the load torque to become greater than the motor
torque, electric drive decelerates and operating point moves away from point B. Similarly when
working at point B and increase in speed will make motor torque greater than the load torque, which
will move the operating point away from point B
 ωm
T

∆ωm TL

Tshift TA TM Torque
From the above discussions, an equilibrium point will be stable when an increase in speed causes load-
torque to exceed the motor torque. (i.e.) When at equilibrium point following conditions is satisfied.
dTL dT
> − − − − − − − − − − − − − (1)
dω m dω m

Basics of Regenerative Braking

In the regenerative braking operation, the motor operates as generator, while it is still
connected to the supply. Here, the motor speed is greater than the synchronous speed. Mechanical
energy is converted into electrical energy, part of which is returned to the supply and rest of the energy
is last as heat in the winding and bearings of electrical machines pass smoothly from motoring region
to generating region, when over driven by the load.

This operation is indicated as shown in the figure below in the first quadrant. Here the power flow is
from the motor to load.

DOWN HILL UPHILL

Speed

Power Flow Power Flow

TL
TM
Speed

M LOAD M LOAD
Speed
TL TM
Torque
Regenerative Braking for DC motor:
In regenerative braking of dc motor, generated energy is supplied to the source. For this the following
condition is to be satisfied.
E > V and Ia should be negative

Speed

Motoring

Braking

Torque

Modes of Operation:
An electrical drive operates in three modes:

Steady state

Acceleration including Starting

Deceleration including Stopping
dω m
We know that T = Tl + J
dt
According to the above expression the steady state operation takes place when motor torque
equals the load torque. The steady state operation for a given speed is realized by adjustment of steady
state motor speed torque curve such that the motor and load torques are equal at this speed. Change in
speed is achieved by varying the steady state motor speed torque curve so that motor torque equals the
load torque at the new desired speed. In the figure shown below when the motor parameters are
adjusted to provide speed torque curve 1, drive runs at the desired speed ω m1 . Speed is changed to ω m 2
when the motor parameters are adjusted to provide speed torque curve 2. When load torque opposes
motion, the motor works as a motor operating in quadrant I or III depending on the direction of
rotation. When the load is active it can reverse its sign and act to assist the motion. Steady state
operation for such a case can be obtained by adding a mechanical brake which will produce a torque in
a direction to oppose the motion. The steady state operation is obtained at a speed for which braking
torque equal the load torque. Drive operates in quadrant II or IV depending upon the rotation.

Tl
ωm

ω m1
1

ωm 2
2

Torque

Acceleration and Deceleration modes are transient modes. Drive operates in acceleration mode
whenever an increase in its speed is required. For this motor speed torque curve must be changed so
that motor torque exceeds the load torque. Time taken for a given change in speed depends on inertia
of motor load system and the amount by which motor torque exceeds the load torque.
Increase in motor torque is accompanied by an increase in motor current. Care must be taken to
restrict the motor current with in a value which is safe for both motor and power modulator. In
applications involving acceleration periods of long duration, current must not be allowed to exceed the
rated value. When acceleration periods are of short duration a current higher than the rated value is
allowed during acceleration. In closed loop drives requiring fast response, motor current may be
intentionally forced to the maximum value in order to achieve high acceleration.
Problems:
A motor having a suitable control circuit develops a torque by the relationship TM = aω + b , where a
and b are positive constants. This motor is used to drive a load whose torque is expressed
as TL = cω 2 + d , where c and d are positive constants. The total inertia of the rotating masses is J.
a) Determine the relations amongst the constants a, b, c and d in order that the motor can
start together with the load and have an equilibrium operating speed?
b) Calculate the equilibrium operating speed?
c) Will the drive be stable at this speed?
d) Determine the initial acceleration of the drive?
e) Determine the maximum acceleration of the drive?
Solution:
a) At ω = 0 , TM=b and TL=d
Hence the motor can start with the load only if b > d
TM=TL at equilibrium speed
i.e. aω + b = cω 2 + d
i.e. cω 2 − aω − (b − d ) = 0
a ± a 2 + 4c(b − d )
Hence ω =
2c
In order that ω is finite a 2 + 4c(b − d ) > 0, which is true
+ Sign before the radical will give a positive ω as long as
a 2 + 4c(b − d ) > 0
− sign before the radial will give a positive ω only if
a a 2 + 4c(b − d )
>
2c 2c
i.e. a > a + 4c(b − d )
2 2

i.e. 4c(b − d ) < 0

i.e c<0, which is not true, since c is given to be a positive constant. Hence the + sign
before the radial only will give a positive finite equilibrium speed.

If a 2 + 4c(b − d ) >0

a + a 2 + 4c(b − d )
b) Equilibrium speed ω =
2c
 dTL dTM
c) = 2cω and =a
dω dω
If the equilibrium speed has to be stable
dTL dTM
> i.e.2cω > a
dω dω
from the answer to (b), we have
2cω = a + a 2 + 4c(b − d ) which will be always > a
Hence, the equilibrium operating speed determined earlier is a stable point of operation
of drive.
dω
d) Accelerating torque J = TM − TL
dt
Initially TM=b and TL=d
b−d
Therefore, initial acceleration =
J
dω
e) Accelerating torque J = TM − TL
dt
= aω − cω 2 + b − d
dω aω − cω 2 + b − d
Therefore, acceleration A = =
dt J
This will be maximum at a speed when
dA
=0
dω
a − 2cω
=0
J
a
ω=
2c
Substituti ng this speed at which the acceleration is maximum,
in the general expression for acceleration, we get

Amax =
(a 2
)
2c − a ( 4c ) + b − d
2

J
a + 4c(b − d )
2
=
4cJ
1. REVIEW OF ROTATIONAL MECHANICS
Rotational mechanics is very important in electric drives studies. The
following Table 1 gives comparison between rotational and linear
mechanics formula.

Table 1 : Rotational mechanics VS linear mechanics.

Rotational Mechanics Linear Mechanics
θ =Angular displacement (rad) S = Displacement (m)

ω = Angular velocity (rad/s) v = velocity (m/s)

α = Angular acceleration (rad/s2) a = Acceleration (m / s2)

J = Moment of inertia (kg.m2) m = Mass (kg)

T = Torque (Nm) = J α F = Force (N) = m a

M = Angular momentum M = Momentum (kg.m /s)

(kg.m2.rad/s) = J ω =m v

Kinetic energy = (Jouls) Kinetic energy = (Jouls)

Power = P = T ω (Watts) Power = F v (Watts)

Work = T θ Work = F S
2. DYNAMICS OF MOTOR- LOAD SYSTEM: FUNDAMENTALS OF
TORQUE EQUATIONS
A motor generally drives a load (Machines) through some mechanical
transmission systems. The equivalent rotational system of motor and load is
shown in the Fig.1.

Fig.1 Typical motor-load system.

where :
J = Moment of inertia of motor load system referred to the motor shaft
kg .m2.
ωm = Instantaneous angular velocity of motor shaft, rad /s.
T = Instantaneous value of developed motor torque, Nm.
Tl = Instantaneous value of load torque, referred to the motor shaft, Nm.

Load torque Tl includes friction and windage torque of motor.

Motor-load system shown in Fig.1 can be described by the following
fundamental torque equation.

( ) ( )

Equation (1.1) is applicable to variable inertia drives such as mine winders,

reel drives, Industrial robots.
For drives with constant inertia

( )

= Torque component called dynamic torque because it is present

only during the transient operations.

At steady state operation .

Note: The energy associated with dynamic torque is stored in the

form of kinetic energy given by

( )
3. Types of Loads
Various load torques can be classified into broad categories, however,
the loads are of two types according to the applied torque:
 Active load torques
 Passive load torques
Load torques which has the potential to drive the motor under equilibrium
conditions are called active load torques. Such load torques usually retain
their sign when the drive rotation is changed (reversed) , for example :
 Torque due to gravitational force.
 Torque due to deformation in elastic bodies and due to tension.
 Torque due to compression and torsion etc.
Load torques which always oppose the motion and change their sign on the
reversal of motion are called passive load torques, for example :
 Torque due to friction.
 Torque due to shear (cutting).
 Torque due to deformation in inelastic bodies etc.
Components of Load Torque (Tl )
The load torque Tl can be further divided into following components :
(i) Friction Torque (TF )
Friction will be present at the motor shaft and also in various parts of the
load. TF is the equivalent value of various friction torques referred to the
motor shaft. Variation of friction torque with speed is shown in Fig 2.

Fig.2 Variation of friction

torque with speed.

Friction torque (TF ) can also be resolved into three components :

(1) VISCOUS friction Tv : Component varies linearly with speed and is

given by

( )
where B is viscous friction coefficient.
 (2) COULOMB friction TC : Component which is independent of
speed.
(3) STATIC friction Ts : Component Ts accounts for additional torque
present at stand still. Since Ts is present only at stand still it is not
taken into account in the dynamic analysis. Its value at stand still is
much higher than its value slightly above zero speed. Friction at zero
speed is called stiction or static friction. In order to start the drive
the motor should at least exceed stiction.

The variation of these three torques are shown in Fig.3. Hence, the total
friction torque is given by

( )

Fig.3 Types of friction torques in drive system.

(ii) Windage Torque (TW )

When motor runs, wind generates a torque opposing the motion. This
is known as windage torque. Windage torque, Tw which is propor-
tional to the square of the speed is given by

( )

(iii) Torque required doing useful mechanical work (TL ).

Nature of this torque depends upon particular application. It may be
constant and independent of speed. It may be some function of
speed, it may be time invariant or time variant, its nature may also
change with the load’s mode of operation.

From the above discussions, and for finite speed,the load torque Tl is

( )
4. BASIC EQUATION OF MOTION FOR DRIVE SYSTEM
Generally, the basic equation of motion of motor driving a load,
Fig.4, is given by

( )

Fig.4 Motor-load system.

where
J = Polar moment of inertia of motor-load system referred to the motor
shaft, Kg.m2
ωm = Instantaneous angular velocity of the motor shaft, rad/sec
Tm = Developed torque of the motor, Nm
TL = Load (resisting) torque, referred to the motor shaft, Nm
TFW = Friction and windage torque =

Coulomb friction is generally neglected in drive systems.

If TFW is small then, TFW = 0 , hence, Eq.(1.8) becomes,

( )

 For motor operation, Tm and ωm have same directions

 Always Tm and TL have opposite directions

In the drive systems, depending on the mechanical load, the motor may
be subjected to variable operating conditions in its duty cycles. The motor in
an electric car can operates in various conditions such as starting, accele-
rating, steady-state, decelerating and stopping. Fig.5 illustrates motor-
load torque characteristics, the available starting torque is Tst. At this
condition, the motor is accelerated and subjected to most severe service.
The equation of motion govern the motor in this case is Eq.(1.8).
 Fig.5 . Motor and load speed-torque characteristics.

 When the motor operates at steady-state:

 When , i.e. the dynamic torque , the drive

accelerating.

 When , i.e. the dynamic torque , the drive

decelerating and coming to rest.

 When , i.e. , the drive continues to run at same

speed if it was running.

 The steady-state operation of the motor occurs when its speed-torque

characteristic intersects with the load speed-torque characteristic at
the operating point A as shown in Fig.5.
Example 1.1

A variable speed d.c. drive has rated power of 10 kW,rated speed of 1500
rpm drives a load that comprises a constant load of TL = 30 Nm. The inertia
of the drive system is 0.10 kg.m2. Calculate the time taken to accelerate the
load from zero to 800 rpm , assuming the drive develops rated torque during
the acceleration phase.

Solution
⁄

( ) ⁄

Example 1.2

An induction motor directly connected to a 400V, 50Hz supply utility has a

rated torque of 30Nm that occurs at a speed of 2940 rpm. The motor drives
a fan load that can be approximated by: TL = B . ωm
where B = 0.05 Nm/rad/s, and the rated speed of the motor is 3000 rpm.
Stating any assumption made, calculate the speed, in equilibrium position at
which the torque developed by the motor is equal to the load torque.
Solution

The torque speed characteristic of an induction motor is shown in Fig.5 ,

part of which can be approximated as straight line.

Let = speed of the motor at full load.

= speed of the motor at no load.

At full load Tm = Trated = 30 Nm

At no-load Tm at = 0 Nm

For the linear region (only)

( )

( )

For the load

For the equilibrium position

( )

⁄
Load Torque and Load Power
At steady state Tm = TL .
The output power from a motor running at speed is

The power required by the load is

If the motor is connected directly to the load as shown in Fig.4, then

Hence (1.10)
If η is the efficiency of the motor on full load, then

( )

In some applications, the motor is connected to the load through a set

of gears. The gears have a teeth ratio and can be treated as speed or
torque transformers. The motor-gear-load connection is shown in
Fig.6.

Fig.6 Motor connected to the load through a gear.

In Fig.6,
Z1 , Z2 = Teeth number in the gear
B1 ,B2 = Bearings and their coefficients
 Jm ,JL = Moment of inertia of the motor and load
The gears can be modelled from the following facts:
(i) The power handled by the gear is the same on both sides.
(ii) Speed on each side is inversely proportional to its tooth number.
Hence
( )

( )
and
( )

Substituting Eq.(1.14) into Eq.(1.13) yields

( )
At steady-state Tm = TL .
If η is the efficiency of the motor on full load, then

( )

and

( )

Now if the motor is geared to the load, then the torque seen by the load is
( )
increased or decreased by the ratio: .
( )

Determination of Referred Load Torque

In the system of Fig. 6 , If the speed of the motor shaft ω1 and that of
the motor is ω2 , T2 is the load torque , T2’ is the load torque referred to
motor shaft , gr = gear ratio = , and ηt = efficiency of transmission,
then equating power :

( )
If the losses in transmission are neglected, then the kinetic energy due to
equivalent inertia is

( )

When there are number (k) of stages of transmission between the driving
motor and the drive load, as shown in Fig.7, Eq.(1.18) becomes:

( )

where : , and are the gear ratios

and the efficiencies of the respective transmission.
Similarly the equivalent inertia will be :

( )

Figure 7 Motor-load system with multi gears.

Referring Forces and Masses Having Translation
Motion to a Rotating One
In some machines or systems, some moving parts rotate while others
undergo translation motion, e.g. cranes, hoists, etc. It is necessary to refer
the translational motion in terms of referred load torque and moment of
inertia referred to motor shaft.
Fig. shows a hoist load lift, wound on drum driven through gears by a
motor. If F is the force required due to gravitational pull to lift the
moving weight W, η is the efficiency of transmission , v (m/s) is the
velocity of the moving mass , and ωm (rad/s) is the angular velocity of the
motor shaft, the referred load torque is obtained by equating the power.

Fig.8 Motor-hoist load system. The

referred load torque TL’:

( )

The moment of inertia referred to the motor shaft is obtained by equating

kinetic energy:

( ) ( ) ( )

where .
Example 1.3
A weight W of (1500 kg) is to be lifted up with a velocity of (1.5 m/s) by
means of a motor-hoist system shown in Figure 8. The winch has a
diameter of (0. 5 m) and driven by motor running at (1000 rpm). The inertia
of the motor and the winch drum are (1.8 kg. ) and (4.2 kg. ) respect-
tively. Calculate the total load torque of the system referred to the motor
shaft .

Solution

Winch drum diameter = o.5 m

Circumference of the winch drum = 0.5π = 1.57 m
Velocity of the weight = 1.5 m /s
Speed of the winch =1.5/1.57 = 0.955 rev/s
ω of the winch, ωwinch = 2π 0.955 = 5.997 rad / s
speed of the motor = 1000/60 = 16.6 rev / s
ωm of the motor = 2π 16.6 = 104.6 rad / s
constant load torque = weight radius of the winch drum
= 1500 (0.5/2) = 375 kgf . m = 375 Nm at ωwinch.
Hence

If is the moment of inertia of the translation movement of the weight

referred to the motor shaft of the motor , then

Total torque referred to the motor shaft :

( ) ( )

( ) ( )

If the system operates at steady-state ( running continuously without

acceleration ) ,then

( )
Example 1.4
A motor has two loads. Load-1 has rotational motion which is coupled to
the motor through a reduction gear with gear ratio gr1 =10 and efficiency of
90%. Load-1 has a moment of inertia of 10 kg.m2 and torque of 10 Nm.
Load-2 has translation motion and consists of 1000 kg weight to be lifted up
at uniform speed of 1.5 m / s. The coupling between load-2 and the motor
has an efficiency of 85%.The motor has an inertia of 0.2 kg.m2 and runs at
constant speed of 1420 rpm. Determine the equivalent inertia referred to the
motor shaft and the power developed by the motor.
Solution
From equations (1.21) :

( )

Jm= 0.2 kg.m2 , gr1 = 10 , v = 1.5 m /s , and

Hence, substituting these values in the above equation we get;

( )

The referred torques are calculated as follows:

The referred torque for load-1 ( ), can be found from Eq.(1.18),

The referred torque for load-2 ( ) can be found from Eq.(1.22),

The total load torque referred o the motor shaft is,

Here: ηt1 = 0.9 , gr1 = 10 , TL1 = 10 Nm , ηt2 = 0.85 , F = mg =1000 9.81

= 9810 N , v = 1.5 m / s , and =148.7 rad / s.
Now substitute these values in the above equation yields;
Example 1.5
A drive used in a hoist to raise and lower weights up to 400 kg at velocities
up to ± 2 m/s. The weight hangs from a cable that is wound on a drum of
radius of 0.4 m . The drum is driven by the drive motor through a gearbox
that has an efficiency of 85% . The maximum speed of the motor is ± 1300
rpm. It is required to:
(a) Sketch the system and find the nearest integer gearbox ratio
that will match the maximum speed of the motor to the maximum
velocity of the hoist.
(b) Determine the torque and power provided by the motor when
lifting the maximum weight at the maximum velocity.
(c ) Calculate the torque and power provided by the motor when
lowering the maximum weight at the maximum velocity.
Solution

(a) The system is shown in Fig. 9.

Figure 9. System diagram of Example 1.5.

The speed of the weight is given as: v = 2 m /s

The motor speed in rad /s is given by

⁄
 (b) Now, when lifting and lowering F is upwards, T1 is in the direction
shown (here anticlockwise)

(Since Force = mass gravity)

When lifting, motor drive supplies the losses in the gearbox :

When lowering , moving mass now supplies gearbox losses, hence

( )

The minus sign indicates that the drive is regenerating.

5. Reducers
The electric motor generally produces relatively high rpm (speed). Since
most of loads in drive systems require low speed operation, therefore it is
required to install a reducer, i.e. an encased transmission mechanism,
between the motor and the actuating mechanism or load. In order obtain
drastic reduction in the speed of motion, the reducers may be fitted with
several series-connected mechanical transmissions. The reducers are mainly
provided by gears, planetary, worm, and screw-gear transmissions. These
types will be discussed briefly hereinafter.
== Gear Transmission
(1) Simple reducers with ordinary gear wheels employ external and
internal gearing (Fig.10)14with efficiency of 98% or more in one pair of
wheels. Such reducers are simple in construction. One pair of gear wheels
is capable of providing small gear ratios ( ) which imply the ratio of
the input speed to the output speed , i.e.:

2
( )
1
 (a) (b)
Fig.10 Single- stage gear transmission with : (a) External transmission,
and (b) Internal transmission.
where Z2 and Z1 are, respectively, the number of teeth on the output and
input gear wheels.
(2) Multi-stage gearing: To obtain greater ratios, multi stage transmission
as shown in Fig.11 , is used. This type is mainly used in low-power drive
systems.The gear ratio of the multi-stage gearing canbe obtained by the
following formula

( )

where n = number of transmission stages ,

= gear ratio of transmission stage.

Fig.11 Multi-stage gear transmission with pairwise meshing.

6. RATING OF MOTORS
Rating or size of motor can be selected in accordance to specific
industrial applications. Beside the rated voltage and rated frequency, the
size of the motor depends also upon:
(1) Temperature rise , which also depend on the duty cycle of the load
 Continuous load
 Intermittent load
 Variable load
(2) Maximum torque required of the motor
Temperature rise :
An electric machine can be considered as a homogeneous body in which
heat is internally developed at uniform rate and heat dissipation is not a rate
proportional to its temperature rise. The relation between the temperature
rise and time is an exponential function which is given by:

( ) ( 6)

where: temperature rise C°

C°

The temperature rise characteristics is depicted in Fig.12.

Fig.12 Temperature rise in electrical machine.

Cooling:
During cooling period (motor speed reduce or stopped) the temperature
equation will be
( 27)
 Example1.6
A motor has a thermal heating time constant of 45 min. When the motor
runs continuously on full rating; its final temperature rise is 75 C°. (a) What
is the temperature rise after two hour if the motor runs continuously on full
load? (b) If the temperature on one hour rating is 70 C°, find the maximum
steady temperature at this rating.
Solution
(a) Heating time constant τ = 45 min.
( ) 75( )
(b)
6
( ) 95.0
736

6.1 Rating of the Motor for Continuous Load

If the motor has load torque T in Nm and it is running at ω rad /sec , the
power rating of the motor:

( 28)

Such loads are pumps, fan, etc.

6.2 Rating of the Motor for Intermittent Loads

Here the motor operating for short time and switch off for long time
(motor is loaded for sometime). The motor is switched on before cooling
completely to the ambient temperature such loads also referred as fluctu-
ating loads, see Fig.13. An approximate and simple method of deter-
mined the rating of a motor subjected to fluctuating load is by assuming that
the heating is proportional to the square of the current drawn by the motor
and hence square of the load. The suitable continuous rating of the motor is
the rms value of the load curve.

Fig.13 Fluctuating loads. e.g. (elevators).

Example10.7

The load cycle of a motor operating an elevator (lift) for 11 minutes is as

follows:

Load period at the bottom 5 minutes 2 hp

Load going up 1 minute 25 hp
Load period at the top 4 minutes 2 hp
Load period going down 1 minute -20 hp

Regenerative braking takes place when the load is disconnected. The cycle
is repeated continuously. Estimate suitable hp for the motor.

Solution

Load variation is plotted in Fig.14.

The total area under the ( ) curve:

A =( ) 5+ 1+ 4+ 1 = 1061 ( ) . min.

The nearest standard rating = 10 hp.

Fig.14 Power variation with time.

6.3 Rating of the Motor for Variable Load
The rating of the motor under such load conditions can be determined
from the load torque vs time carve as depicted in Fig.15. This is called the
method equivalent torque. In case of machines, whose flux remains
constant irrespective of load variation, the equivalent torque rating is given
by:

∑ ( 29)
√
∑

Fig.15 Load torque variation with time.

If the speed at which the load operates is approximately constant, the power
P is proportional to the torque T and,

∑
√ ( 0)
∑
Example 1.8

An electric motor has load variation as:

Torque 250 Nm for 25 minutes
150 Nm for 10 minutes
320 Nm for 12 minutes
180 Nm for 20 minutes
Find the equivalent torque rating of the motor. If the speed of the motor is
1000 rpm find the power rating of the motor?
Solution
( ) ( ) ( ) ( )

Power rating of the motor

Example 1.9

A motor driving an industrial load which follows the following cycles:

Power 50 kW for 15 minutes
No load for 5 minutes
30 kW for 10 minutes
No load for 8 minutes
The cycle repeated indefinitely. Find the suitable size of the continuously
rated motor for the purpose.

Solution
( ) ( )

The nearest standard motor size is 37 .

 CONVERTER / CHOPPER FED DC DRIVES

2.1 INTRODUCTION

Direct-current motors are extensively used in variable-speed drives and position-

control systems where good dynamic response and steady-state performance are
required. Examples are in robotic drives, printers, machine tools, process rolling mills,
paper and textile industries, and many others. Control of a dc motor, especially of the
separately excited type, is very straightforward, mainly because of the incorporation of
the commutator within the motor. The commutator brush allows the motor-developed
torque to be proportional to the armature current if the field current is held constant.
Classical control theories are then easily applied to the design of the torque and other
control loops of a drive system.
2.2 DC MOTORS AND ITS CHARACTERISTICS
When a DC supply is applied to the armature of the dc motor with its field excited
by a dc supply, torque is developed in the armature due to interaction between the axial
current carrying conductors on the rotor and the radial magnetic flux produced by the
stator. If the voltage V is the voltage applied to the armature terminals, and E is the
internally developed motional e.m.f. The resistance and inductance of the complete
armature are represented by Ra and La in Figure 2.1(a). Under motoring conditions, the
motional e.m.f. E always opposes the applied voltage V, and for this reason it is referred
to as ‘back e.m.f.’ For current to be forced into the motor, V must be greater than E, the
armature circuit voltage equation being given by
dI a
V = E + I a Ra + L (2.1)
dt
The last term in equation (2.1) represents the inductive volt-drop due to the armature self-
inductance. This voltage is proportional to the rate of change of current, so under steady-
state conditions (when the current is constant), the term will be zero and can be ignored.
Under steady-state conditions, the armature current I is constant and equation (2.1)
simplifies to
V = E + I a Ra ( 2 .2 )
 2.2.1 Types of DC Motors
Based on the connections of armature and field windings DC
motors classified in to three types they are
a. Separately excited DC motor [Field and armature windings are excited
separately by independent sources fig2.1(a) ]
b. Shunt excited DC motor [ Here field winding and armature winding are
connected in parallel and are excited by a common source fig 2.1 (b) ]
c. Series excited DC motor [ Here field winding and armature winding are
connected in series and are excited by a common source fig 2.1 (c) ]

Fig 2.1 (a) Fig 2.1 (b) Fig 2.1 (c)

2.2.2 Speed -Torque and Speed -Current Relations
The motor back emf is given by
φZN  P 
Eb =  Volts (2.3)
60  A 
Where φ is flux per pole in Webers
Z is number of armature conductors
N is speed in rpm
P is number of parallel paths in armature
Here Z, P, A are fixed for a particular machine after wounded. Therefore for a given DC
machine
 ZP 
Eb =  φN Volts (2.4)
 60 A 
E b = K aφN (2.5)

60
Where N = ω m , substitute in equation (2.4)
2π
 ZP   60 
Eb =  φ  ω m Volts
 60 A   2π 
 ZP 
Eb =  φω m
 2πA 
E b = K aφω m (2.6)
The torque developed by the armature is given by
 φZI a
P
Ta =  N − m (2.7)
2π  A
 ZP 
= φI a
 2πA 
Ta = K aφI a (2.8)

ZP
Where K a =
2πA
Ta = K aφI a = T (2.9)

2.2.2.1 Separately Excited or Shunt Motor

From expression (2.2) I a =

(V − E ) (2.10)
Ra
Substituting equation (2.10) in equation (2.6) we get
V − E 
Ta = K aφ   (2.11)
 Ra 
Substituting equation (2.9) in equation (2.11) we get
V − K aφω m 
Ta = K aφ   (2.12)
 Ra 
Rearranging the above equation we get,
V Ra
ωm = − Ta (2.13)
K a φ ( K a φ )2

The above expression gives the relationship between speed and torque for
separately excited and shunt motors.
Ta
Speed –Current relationship can be obtained if in the expression (2.13) is
K aφ
replaced with Ia (From equation 2.9) as given below
V R I
ωm = − a a (2.14)
K aφ (K a φ )
Fig 2.2 (a) and 2.2 (b) shows the speed torque characteristics and Speed current
characteristics of separately excited and shunt motor when the armature and field
voltages are kept constant.
 Fig 2.2(a) Fig 2.2 (b)
Speed-Torque Characteristics Speed-Current Characteristics
2.2.2.2 Series Motor
In the case of separately excited and shunt motor flux is almost constant when the
armature voltage is fixed, but in case of series motor when the machine is loaded
armature current increase which increase the flux also, because armature and field
windings are in series in the case of series motor. Assuming that the motor operates in the
linear region of the magnetic saturation curve we get,
φ = CI a (2.15)
Where C is proportionality constant. The torque developed in the armature in this case is
given by, i.e equation (2.9) can be rewritten as
Ta = K aφI a = K a CI a I a = K a CI a
2
(2.16)
Therefore equation (2.13) and (2.14) becomes
V Ra
ωm = − Ta (2.17)
K a CI a (K a CI a )2

V Ra
ωm = − (2.18)
K a φ (K a C )
The Speed torque characteristic of DC series motor is shown in the figure 2.3.
Note that the speed of the motor is rapidly decreasing when the load is increased.

Fig 2.3

Speed Torque Characteristic of DC series motor.

2.3 Conventional Methods of Speed Control of DC Motors
2.3.1 Speed control of separately excited or DC shunt motor
As seen from in the above section 2.2, the speed torque characteristic and speed current
characteristic of separately excited or shunt motor is as given below,
V Ra
ωm = − Ta = ω 0 - ∆ω (2.19)
K a φ ( K a φ )2

V R I
ωm = − a a = ω 0 - ∆ω (2.20)
K aφ (K a φ )

Where ω 0 the no load is speed and ∆ω is the speed drop. The no load speed is computed
when the torque and current are equal to zero. The speed drop is a function of the load
torque. From the above expressions speed of the separately excited DC motor or Shunt
motor can be controlled by controlling the following quantities:
a) Resistance in the armature circuit: When the resistance is inserted in the
armature circuit, the speed drop ∆ω increases and the motor speed decreases.
b) Terminal Voltage (Armature voltage): Reducing the armature voltage V of the
motor reduces the motor speed.
c) Field Flux (or Field Voltage): Reducing Field voltage reduces the flux φ , and
the motor speed increases.
Note: We cannot operate the electric motor with voltages higher than the rated value.
Therefore we cannot control the speed by increasing the armature or field voltages
beyond the rated values. Only voltage reduction can be implemented. Hence second
method of speed control is only suitable for speed reduction (armature voltage), where
third method (Field flux) is suitable for speed increase.
2.3.2 Controlling speed by adding external resistance to armature.
Figure 2.4 shows a DC motor setup with resistance added in the armature circuit.
Figure 2.5 shows the corresponding speed torque characteristics. Let us assume that the
load torque is unidirectional and constant. A good example for this type of torque is
elevator. Also assume that the field and armature voltages are constant.

Fig 2.4 Fig 2.5

A setup for speed change by adding Effect of adding an armature resistance
an armature resistance on speed
At point 1 no external resistance is added in the armature circuit. If a resistance
Radd1is added to the armature circuit, the motor operates at point 2, where the motor speed
ω 2 is
V R + Radd 1
ω2 = − a T = ω − ∆ω 2 (2.21)
K aφ ( Kφ ) 2 a 0
or
V R + Radd 1
ω2 = − a I a = ω 0 − ∆ω 2 (2.22)
K aφ K aφ

Note that the no load speed ω 0 is unchanged regardless of the value of resistance

in the armature circuit. The second term of the speed equation is the speed droop ∆ω ,
which increase in magnitude when Radd increases. Consequently, the motor speed is
reduced. If the added resistance keeps increasing, the motor speed decreases until the
system operates at point 4, where the speed of the motor is zero. The operation of the
drive system at point 4 is known as “holding”.
Note: Operating a dc motor for a period of time with a resistance inserted in the
armature circuit is a very inefficient method. The use of resistance is acceptable only
when the heat produced by the resistance is utilized as a by product or when the
resistance is used for a very short period of time.
2.3.2 Controlling speed by adjusting armature voltage.
A common method of controlling speed is to adjust the armature voltage. This
method is highly efficient and stable and is simple to implement. The circuit of figure 2.6
shows the basic concept of this method.

Fig 2.6
A setup for speed change by adjusting armature voltage
The only controlled variable is the armature voltage of the motor, which is
represented as an adjustable voltage source. Based on equation 2.19, when armature
voltage is reduced no load speed is also reduced. Moreover for the same value of load
torque and field flux, the armature voltage does not affect the speed drop ∆ω . The slope
Ra
of speed torque characteristic is , which is independent of the armature voltage.
( K a φ )2
Hence the characteristics are shown as figure 2.7. Note that it is assumed that the field
voltage is unchanged when the armature voltage varies.

Fig 2.7
2.3.2 Controlling speed by adjusting Field voltage
Equations (2.19) and (2.20) show the dependency of motor speed on the field
flux. The no load speed is inversely proportional to the flux and slope of equation (2.19)
is inversely proportional to square of the flux. Therefore the speed is more sensitive to
flux variations than to variations in the armature voltage.
Figure 2.8(a) shows a setup for controlling speed by adjusting the field flux. If we
reduce the field voltage, the field current and consequently the flux are reduced. Figure
2.8 (b) shows a set of speed torque characteristics for three values of field voltages. When
the field flux is reduced, the no load speed ω 0 is increased in inverse proportion to the

flux, and the speed drop ∆ω is also increased.

Fig 2.8 (a) Fig 2.8(b)

The characteristics show that because of the change in speed drops, the lines are
not parallel. Unless the motor is excessively loaded, the motor speed increases when the
field is reduced. When motor speed is controlled by adjusting the field current, the
following considerations should be kept in mind:
a. The field voltage must not exceed the absolute maximum rating
b. Since DC motors are relatively sensitive to variations in field voltage, large
reductions in field current may result in excessive speed.
c. Because the armature current is inversely proportional to the field

flux  I a =  , reducing the field results in an increase in the armature

Ta
 K aφ 

current (Assuming that the load torque is unchanged).

Thus by combining armature and field control for speeds below and above rated
speed, relatively a wide range of speed control is possible. For speeds lower than
that of the rated speed, applied armature voltage is varied while the field current is
kept at its rated value: to obtain speeds above the rated speed, field current is
decreased while keeping the applied armature voltage constant.
 Examples of Speed Control of Separately and shunt d.c. Motors

Example 1

A separately-excited d.c. motor used to drive a fan whose torque is

proportional to the square of the speed. When the armature circuit of the
motor is connected across 200 V, it takes armature current of 16 A and the
motor runs at speed of 1000 rpm. If the speed of the motor is to be
reduced to 750 rpm, calculate the required terminal voltage and the
current drawn by the motor at the new speed. Assume the armature
resistance is 0.5 Ω and neglect brush voltage drop.

Solution
Armature current = 16 A

Back emf
Speed n1 =1000 rpm

Torque

Let the voltage required to lower the speed to 750 rpm be

,
To find we must find and
To find : since for a separately-excited motor , the flux is constant,

Also

To find :
Example 2

A 240 V d.c. shunt motor has an armature resistance of 0.2 Ω. When the
armature current is 40 A, the speed is 1000 rpm. (a) Find additional
resistance Rx to be connected in series with armature to reduce the speed
to 600 rpm. Assume the armature current remains the same. (b) If the
current decreases to 20 A (with resistance Rx connected) find the new
speed of the motor.

Solution
(a)

From which :

(b)

From which :
Example 3

A 300 V d.c. shunt motor runs at 1600 rpm when taking an armature
current of 40 A. The armature resistance is 0.5 Ω. It is required to :
(a) Calculate the speed when a resistance is inserted in the field
circuit as to reduce the flux to 80% of its nominal value (flux
weakening),
(b) Calculate the speed when the field resistance is deacrease to a
value such that the flux is increasead to 120 % of its nominal
value .
Assume that the armature current remains constant in both cases.

Solution
(a) In speed control of d.c. motor using flux variation method, the
terminal voltage and armature resistance are kept constant,while the
flux is varied. Therefore,

By dividing equation (a) by (b) yields,

Since , hence

(b) When the flux is increased by 120%

Note : Higher speeds can be obtained with field weakening control

method.
 Example of Speed Control of Series d.c. Motors
Example 11.4

A d.c. series motor draws 22 A from a 240 V line while running at 840
rpm. It has an armature resistance of 0.6 Ω and a series field resistance of
0.5 Ω. A diverter is to be added to the circuit so that the speed increases to
1200 rpm while the line current increases to 28 A. Find the value of the
diverter resistance Rd .
Solution

Now using Eq.(11.38)

From which
FOUR-QUADRANT OPERATION OF A DRIVE SYSTEM AND MOTOR BRAKING

FOUR-QUADRANT OPERATION OF DC MACHINE

A d.c. machine can operate as a motor, as a generator or as a brake as
illustrated in the following diagram (Fig.1 ). It has been assumed in
this diagram that the field current is fixed (in magnitude and direction)
and the armature reaction is negligible such that Kφ is constant. In this
case the ω –T equation is linear.

Assumptions :
 The positive or forward speed is arbitrarily chosen in counterclo-
ckwise direction (it can also be chosen as clockwise). The positive
torque is in the direction that will produce acceleration in forward
speed, as shown in Fig.1.
 The plane of Fig.1 is divided into four quadrants, thus four
modes of operation. The quadrants are marked as: I, II, III and IV.

Fig.1 Four-quadrant operation of a d.c. motor.

Quadrant I
Both torque and speed are positive – the motor rotates in forward
direction, which is in the same direction as the motor torque. The power
of the motor is the product of the speed and torque (P = T e ω), therefore
the power of the motor is positive. Energy is converted from electrical
form to mechanical form, which is used to rotate the motor. The mode of
operation is known as forward motoring.

Quadrant II
The speed is in forward direction but the motor torque is in opposite
direction or negative value. The torque produced by the motor is used to
‘brake’ the forward rotation of the motor. The mechanical energy during
the braking is converted to electrical energy – thus the flow of energy is
from the mechanical system to the electrical system. The product of the
torque and speed is negative thus the power is negative, implying that the
 motor operates in braking mode. The mode of operation is known as
forward braking.

Quadrant III
The speed and the torque of the motor are in the same direction but are
both negative. The reverse electrical torque is used to rotate the motor in
reverse direction. The power, i.e. the product of the torque and speed, is
positive implying that the motor operates in motoring mode. The energy
is converted from electrical form to mechanical form. This mode of
operation is known as reverse motoring.

Quadrant IV
The speed is in reverse direction but the torque is positive. The motor
torque is used to „brake‟ the reverse rotation of the motor. The mechanical
energy gained during the braking is converted to electrical form, thus
power flow from the mechanical system to the electrical system. The
product of the speed and torque is negative implying that the motor
operates in braking mode. This mode of operation is known as reverse
braking.

ELECTRICAL BRAKING OF DC MOTORS

A motor is said to be in braking mode when Te and ωm ( motor speed)

are in opposite direction (Figure . 2). If Ea becomes ˃ VT for any
reasons, then Ia will become negative (reverse) and Te will become in the
same direction of TL ; which opposes rotation. Hence the speed will
reduced (since negative dynamic torque is acting on the motor shaft) as
described in the following equations.

Fig.2.

Te remain negative until Tj = 0.

 If the motor supply is disconnected while motor speed is ω1 then it
takes some time, tstop , until it reaches zero speed. In this case the
developed motor speed is zero and the accelerated torque Tj is negative
(acting to decelerate the motor).

The stop time , tstop , can be determined from

∫ ∫

During the deceleration period, the stored energy is completely

consumed in supplying rotational losses and in supplying the coupled load
by the required mechanical energy as input. Finally, the speed of the
rotating part (the rotor of the motor and its coupled load) attains zero
whereby the stored kinetic energy is zero. Note that TL and Tloss are, in
general , rising functions of speed and being very small at low speeds, and
therefore tstop is high.

Types of d.c. Motor Electric Braking

There are three types of electrical braking applied to the d.c. motors,
namely, regenerative braking, plugging braking and dynamic breaking.

1. Regenerative Braking
 This type of braking happen when the motor speed increases above
the no-load speed ωmo (for example, lowering of a load by
electrically operated winch and when an electric train goes
downhill). The mechanical energy in this type of breaking is
converting into electrical energy, part of which is return to the
supply and the rest of the energy is lost in the machine.
 Most of the motors pass smoothly from motoring to generating
operation if the induced emf Ea exceeds the source voltage Vt (due
to increase of motor speed from ωmp to ωmr ). In this case the
current Ia becomes negative and the machine will act as a generator
pumping power back into the source. This regeneration created by
the negative TL which accelerates the machine from point 1 to
point 2 , picking up the speed in excess of the no load speed ω mo at
point 3 as shown in Figure. 3. Under this condition Ea = K ϕ ωmr
(ωmr ˃ ωmo) becomes greater than the supply voltage Vt.
 ωm
2 3
ωmr

ωmo
1
ωmo
ω mp
Vt
G M

0 TL Ia, Te

Figure 3. Regenerative braking

Consider now a separately-excited d.c. motor in regenerative mode.

Figure. 4(a) shows the motor working at is normal state at point 2.
Now for the transition from 1 to 2 in general case, for Figure.4 (b),

5
The equivalent diagram for this transision is depicted in Figure. 4.
Since in regenerative breaking : ωmr ˃ ωmo , then at point 2

Figure. 4: Separately-excited d.c. motor in regenerative mode.

To find the value of the electromagnetic torque developed by the motor in
the regenerative braking condition, substitute Eq.(6) into the follo-
wing torque equation yields

Equation (7) indicates that, in regenerative breaking condition, the

motor torque has negative value.
To find an analytical expression for the mechanical characteristic in
regenerative condition, equation (7 ) must be solved with respect to
, that is

where is the no-load ideal speed.

Notes :
 To maintain the current below the maximum permissible value, an
external resistance Rx may be needed for this purpose.
 The series motor cannot be used in the regenerative breaking
condition, i.e at negative load torque (-Tm).

2. Plugging

This type of breaking is applicable for all types of d.c. motors, namely
separately-excited, shunt, compound and series motors. In this method the
polarity of the applied terminal voltage of the motor is reversed. As a
result the motor torque Tm reverses its direction and acts as a break to the
motor shaft by reducing its speed to zero. At this instant, i.e. when
, the supply must be switched off otherwise the motor will run in
reverse direction with negative speed. Figure. 5 shows braking of
separately-excited motor by plugging method.
It is important that, during voltage reversal an external resistance Rx
should be inserted with the armature circuit to limit the braking current.
Referring to Figure. 6 the sequences of events during plugging for
separately excited motor are :
From point 1 to point 2 : Current and torque reversal

9
 PLUGGING MOTORING

Fig.5 Motor braking by plugging.

Fig.6 Plot of mechanical characteristics of separately-excited motor

in plugging mode.

At point 3 :

The motor start to run in opposite direction toward point 4.

3. Dynamic Braking

This method of braking the motor is disconnected from the supply and
operated as a generator by the kinetic energy of rotor. The kinetic energy
is then dissipated in an external resistance connected across the motor.
With this technique, the energy required from the supply to brake the
motor is eliminated as compared to the previous plugging method. This
method of braking can be applied to brake d.c. motors, synchronous a.c.
motors as well as a.c. induction motors and generally referred to as
Rheostatic Braking. See Fig.7.

MOTORING BRAKING

Fig.7 Dynamic breaking of d.c. separately-excited motor.

The sequences of events during plugging are:

From 1 to 2 : Current reversal

Note:
 Short circuiting the motor makes Vt = 0.
 Rx is used to limit the current and to dissipate the stored kinetic
energy.
 Example 5

A 30 kW, 415 V d.c. shunt motor is braked by plugging. Calculate the

value of the external resistance Rx to be placed in series with the armature
circuit to limit braking current to 164 A. The armature resistance of the
motor is 0.1 Ω and the full load armature current is 100 A at full load
speed of 600 rpm. What is the braking torque obtained from the motor?
Solution

The emf induced in the motor

Voltage across the armature at the instant of braking

The total resistance required to limit the current to 164 A

Full load torque :

Since the flux in the shunt motor is constant, hence the torque is
proportional to the armature current, hence the initial braking torque is
Example 6

A 600 V d.c. shunt motor having an efficiency of 80% operates a hoist

having an efficiency of 75% . Determine the current taken from the
supply when the hoist raise the load of 450 kg at speed of 3 m/s. If
rheostat braking is used to brake the motor, what is the value of the
external resistance must be put in series with the armature circuit in order
to lower the same load at the same speed?

Solution
Work done when raising the load

Current drawn from the supply

When the load is lowered, the motor will operate as a generator. The
output of the generator will be dissipated in the resistance. Assume the
generator has efficiency of 85%, hence,

If we neglect the armature resistance, the inserted external resistance is

 Lecture Five
In this lecture we shall consider mainly the DC motors spead control
using armature voltage control by AC-to-DC converters .These single-
phase and three-phase converters (drives) are shown in Figures
below
 Dual converters
The dual converter consists of two AC-to-DC converters connected in
anti-parallel as shown in Figure. If converter 1 operates, and it gives a
positive output dc voltage (+ Vdc) .However if converter2 operates it
gives negative dc voltage -Vdc. It is important to be noted that, the two
converters must not operate simultaneously to avoid short - circuiting of
the two converters.

The dual converter provides the facility of operation in four quadrants.

If the load is a dc motor, the motor can runs in four modes of operation
as indicated in Figures.
Vdc

Q2 Q1
Idc
Q3 Q4
2.4 Controlled Rectifier Fed DC Drives
Controlled rectifiers are used to get variable dc voltage from an ac source of fixed
voltage. There are several types of converters which can be used for feeding DC motors.
AS thyristors are capable of conducting current in one direction all theses rectifiers are
capable of conducting current only in one direction.
2.4.1 Types of Rectifiers
AC to DC converters
(Or) Rectifiers

Single Phase type Three phase type

Uncontrolled Controlled
(Contains only DIODES) (Contains both SCR and Diodes)

Half Wave Full Wave

(1 Pulse) (2 Pulse)

Half wave full Half wave semi Full wave full Full wave semi
Converters without Converter with Converters without Converter with
Freewheeling Diode Freewheeling Diode FWD FWD

2.4.2 Single Phase rectifier fed separately Excited DC motor drive

For motors up to a few kilowatts the armature converter can be supplied from either
single-phase or three-phase mains, but for larger motors three-phase is always used. A
separate thyristor or diode rectifier is used to supply the field of the motor: the power is
much less than the armature power, so the supply is often single-phase. Figure 2.9 shows
the setup for single phase controlled rectifier fed separately excited dc motor drive.
Field circuit is also excited by a dc source, which is not shown in the figure just for
simplicity.

Fig 2.9
 The motor terminal voltage waveform and current waveform for the dominant
discontinuous and continuous conduction modes are shown in the figure 2.10(a) and 2.10
(b).

Fig 2.10 (a) Discontinuous Conduction Mode Waveforms

Fig 2.10 (b) Continuous Conduction Waveforms
 a a
2.4.2.1Discontinuous Conduction:

Va = E + I a Ra (2.23)
From discontinuous waveforms

1 
β π +aα
Va =  ∫ Vm sin ωt d (ωt ) + ∫ E d (ωt )
π α β 

=
Vm
[cos α − cos β ] + (π + α − β )E [Q ∫ sin θdθ = cos θ ] (2.24)
π π

From equations (2.23) and (2.24)

Vm  β −α 
[cos α − cos β ] − I a Ra
= E  (2.25)
π  π 
Substituting (2.6) and (2.8) in (2.25)
Vm  β −α 
[cos α − cos β ] − T
Ra = K ω m   [Where K = K a φ ]
π K  π 
 β − α  Vm
ωm  = [cos α − cos β ] − T2 Ra
 π  Kπ K

ωm =
Vm
[cos α − cos β ] − T2 .Ra . π
K (β − α ) K (β − α )
Vm  cos α − cos β  Ra π
ωm =   − 2 T (2.26)
K  ( β) − α  K (β − α )
For a given α , there is a particular speed ω mc when β = π + α , indicating that at ω mc , the

mode of operation changes from discontinuous to continuous.ω mc is called as critical

speed. Substituting β = π + α in equation (2.23) we get,

Ra 1 + e −π cot φ 
ω mc = Vm sin (α − φ ) −π cot φ  (2.27)
KZ e − 1
Continuous Conduction Mode
For continuous conduction, average output voltage is given by,
π +α
1
Va =
π ∫ Vm sin ωtd (ωt )
α
(2.28)

2Vm
Va = cos α (2.29)
π
2Vm R
ωM = cos α − a2 T (2.30)
πK K
2.4.3 Three Phase Fully Controlled rectifier fed separately Excited DC motor drive

Three phase controlled rectifiers are used in large power DC motor drives. Three
phase controlled rectifier gives more number of voltage pulses per cycle of supply
frequency. This makes motor current continuous and filter requirement also less

The number of voltage pulses per cycle depends upon the number of thyristors
and their connections for three phase controlled rectifiers. In three phase drives,
the armature circuit is connected to the output of a three phase controlled rectifier.

Three phase drives are used for high power applications up to mega watts power
level. The ripple frequency of the armature voltage is higher than that of the
single phase drives and it requires less inductance in the armature circuit to reduce
the armature current ripple.

Three phase full converters are used in industrial applications up to 1500KW
drives. It is a two quadrant converter. i.e. the average output voltage is either
positive or negative but average output current is always positive.

2.4.3.1 Principle of Operation:

Three phase full converter bridge circuit connected across the armature terminals
is shown in the figure 2.11 and figure 2.12 shows the voltage and current waveforms
of the converter. The circuit works as a three phase AC to DC converter for firing
angle delay 0 0 < α < 90 0 and as a line commutated inverter for 90 0 < α < 180 0 . A
three phase full converter fed DC motor is performed where regeneration of power is
required i.e. it performs two quadrant operation.

Figure 2.11
Basically, the controlled rectifier consists of six thyristors arranged in the form of
three legs with two series thyristors in each leg. The center points of three legs are
connected to a three-phase power supply. The transformer is not mandatory, but it
provides the advantages of voltage level change, electrical isolation, and phase shift from
the primary. In a three-phase bridge, one device in the positive group (Q1 Q3 Q5) and
another device from the negative group (Q4 Q6 Q2) must conduct simultaneously to
contribute load current id. Each thyristor is normally provided with pulse train firing for
the desired conduction interval. The speed of the motor can be controlled by firing angle
control of the thyristors.

Fig 2.12 Three-phase thyristor bridge waveforms in rectification mode ( α = 40°)

 Fig 2.13 Three-phase thyristor bridge waveforms inverting mode ( α = 150°)
The average motor armature voltage is given by

π
+α
2
3
Va =
π π
∫V ab d (ωt ) (2.31)
+α
6

 π
In the above substitute Vab = 3Vm sin  ωt + d (ωt ) (2.32)
 6
3 3
We have Va = Vm cos α (2.33)
π
2.4.3.2 Speed Torque Relations:
The drive speed is given by
Va = E b + I a Ra Where E b = K aφω

Then Va = K aφω m + I a Ra

Va − I a R a
ωm = (2.34)
K aφ

In separately excited DC motor K aφ I a = T therefore (2.34) becomes

Va Ra
ωm = - T (2.35)
K aφ (K a φ )2
2.5 Chopper Fed DC drives
 A chopper is a static device that converts fixed DC input voltage to a variable dc
output voltage directly
 A chopper is a high speed on/off semiconductor switch which connects source to
load and disconnects the load from source at a fast speed.
 Choppers are used to get variable dc voltage from a dc source of fixed voltage.
Self commutated devices such as MOSFET’s, Power transistors, IGBT’s, GTO’s
and IGCT’s are used for building choppers because they can be commutated by a
low power control signal and do not need communication circuit and can be
operated at a higher frequency for the same rating.
 Chopper circuits are used to control both separately excited and Series circuits.

2.5.1 Advantages of Chopper Circuits

Chopper circuits have several advantages over phase controlled converters
1. Ripple content in the output is small. Peak/average and rms/average current ratios
are small. This improves the commutation and decreases the harmonic heating of
the motor.
2. The chopper is supplied from a constant dc voltage using batteries. The problem
of power factor does not occur at all. The conventional phase control method
suffers from a poor power factor as the angle is delayed.
3. Current drawn by the chopper is smaller than in phase controlled converters.
4. Chopper circuit is simple and can be modified to provide regeneration and the
control is also simple.
2.5.2 Chopper Controlled Separately Excited DC motor
If the source of supply is D.C. (for example in a battery vehicle or a rapid transit
system) a chopper-type converter is usually employed. The chopper-fed motor is, if
anything, rather better than the phase-controlled, because the armature current ripple can
be less if a high chopping frequency is used.
2.5.2.1 Motoring Mode of Operation
A transistor is used to chop the DC input voltage in to pieces and chopped DC
voltage is given to the motor as shown in the figure 2.14. Current limit control is used in
chopper. In current limit control, the load current is allowed to vary between two given
limits (i.e. Upper and lower limits). The ON and OFF times of the transistor is adjusted
automatically, when the current increases beyond the upper limit the chopper is turned
off, the load current free wheels and starts to decrease. When the current falls below the
lower limit the chopper is turned ON. The current starts increasing if the load. The load
current and voltage waveforms are shown in the figure 2.15. By assuming proper limits
of current, the amplitude of ripple can be controlled.

Fig 2.14 Fig 2.15

The lower the current ripple, the higher the chopper frequency. By this switching
losses increase. Discontinuous conduction avoid in this case. The current limit control is
superior one.
Duty Interval
During the ON period of the chopper (i.e) duty interval 0 <t<tON, motor terminal voltage
Va is a source voltage V and armature current increases from ia1 to ia2. The operation is
describe by,
dia
R a I a + La + E = V 0 ≤ t ≤ t ON (2.36)
dt
In this interval the armature current increases from Ia1 to Ia2 since the motor is connected
to the source during this interval, it is called as duty cycle.
Free Wheeling Interval
Chopper Tr is turned off at t=tON. Motor current free wheels through the diode D
and the motor terminal voltage is zero. During interval t ON ≤ t ≤ T . Motor operation
during this interval is known as free wheeling interval and is described by
di a
R a I a + La + E = 0 t ON ≤ t ≤ T (2.37)
dt
During this interval current decreases from ia2 to ia1
Duty cycle (or) Duty Ratio:
Duty cycle is defined as the ratio of duty interval tON to chopper period T is called Duty
cycle (or) Duty Ratio.
Duty Interval t
δ= = ON (2.38)
Chopper Period T
From figure 2.15
t
1 ON
Va = ∫ Vdt (2.39)
T 0

Solving the above,

∫ dt = T [t ]
tON
V V t ON tON
Va = =V (2.40)
T 0
0 T
Va = δV (2.41)
Then the speed of the chopper drive can be obtained as
Va = E + I a R a
Substituting Va from equation (2.41) in the above equation we get,
δV = E + I a Ra (2.42)

Substituting E = Kω m we get
 δV − Kω m
Ia = (2.43)
Ra
From above equation we get

δV I a Ra
ωm = − (2.44)
K K
Substituting T = KφI a in above equation we get

δV Ra T (2.45)
ωm = −φ2
K K
The torque speed characteristics of chopper fed separately excited DC motor is shown in
the figure 2.16

Fig 2.16
2.5.2.2 Regenerative Braking Mode
Regenerative braking operation by chopper is shown in the figure 2.17. Regenerative
braking of a separately excited motor is fairly simple and can be carried out down to very
low speeds.

Fig 2.17 Fig 2.18

In regenerative mode, the energy of the load is fed back to the supply system. The
DC motor works as a generator during this mode. As long as the chopper is ON the
mechanical energy is converted in to electrical energy by the motor, now working as a
generator, increases the stored magnetic energy in the armature circuit. When chopper is
switched off, a large voltage appears across the motor terminals this voltage is more than
that of the supply voltage V and the energy stored in the inductance and energy supplied
by the machine is fed back to the supply system. When the voltage of the motor fall to V,
the diodes in the line blocks the current flow preventing any short circuit of the load can
be supplied to the source. Very effective braking of motor is possible up to extreme small
speeds.
Energy Storage Interval
The stored energy and energy supplied by the machine is fed to the source. The
interval 0 <t<tON is now called energy storage interval and interval t ON ≤ t ≤ T is the duty
interval.
T − tON
Here duty ratio δ = (2.46)
T
From figure 2.18

∫ dt = T [t]
T T
1 V V V
(T - t ON )
T
Va = ∫
T tON
Vdt =
T t ON
t ON
=
T
(2.47)

 T − t ON   t 
Va = V   = V 1 − ON  (2.48)
 T   T 
Therefore the speed torque relations under braking operation is given as

ωm =
(1 − δ )V −
Ra
T (2.49)
K K 2φ
2.5.3 Chopper control of DC series motor

The speed of this drive ω m can be derived from the following equation

E = Va - I a Ra but Va = δV
∴ E = δV - I a Ra K aωm = δV - I a Ra
δV - I a Ra
ωm =
Ka