i.

Dynamics of Electrical drives

• A motor generally drives a load ( machine) through some

transmission system.
• While motor always rotates, the load may rotate or undergo
translational motion, or both simultaneously.
• Load speed may be different from that of the motor.
• Representation of motor-load system may seem the following.

Motor and load torque representation

• Motor-load system can be described by the following
fundamental torque equation (Equation of motion).

d m
T  TL   J m   J  m
d dJ
dt dt dt

d m
For drives with constant inertia, (dJ/dt) = 0; thus,

T  TL  J
dt
Where, T = developed motor torque
TL = Load torque referred to motor shaft.
J = polar moment of inertia of motor load system referred to motor shaft
ωm = angular velocity of motor shaft
This equation shows that, torque developed by the motor is Counter
balanced by a load torque TL and a dynamic torque J(dω/dt).
• Torque component J(dω/dt) is called the dynamic torque
because it appears during the transient operation.
• Moment of inertia of a drive may be expressed as;

J  m  kg  m 2
GD 2
2

4g
Where, ρ and D – radius and diameter of gyration, respectively
G – weight, N
g – 9.81 m/c2 acceleration due to gravity
Analysis of the fundamental torque equation shows that;

 When T > TL, dω/dt > 0. i.e. the drive undergo acceleration.
 When T < TL, dω/dt < 0. i.e. the drive will undergo deceleration.
Deceleration occurs also at negative values of motor torque ( i.e. during
braking operation).
 When T = TL, dω/dt = 0, i.e the drive will run at a steady state speed.
Speed –torque conventions and multi-quadrant
operations.

• For considerations of multi-quadrant operation of drives, let us

see suitable conventions about the signs of torques and speed.

• Motor speed is considered positive when rotating in the forward

direction. For drives which operate only in one direction, forward
speed will be their normal speed.

• In loads involving up-and-down motions, the speed of motor

which causes upward motion is considered forward motion.
• For reversible drives, forward speed is chosen arbitrarily. Then the
opposite direction gives reverse speed.

• Motor torque is considered positive, when it produces

acceleration (positive rate of change of speed) in forward direction.
• Positive load torque is opposite in direction to the positive motor
torque.
• Motor torque is considered negative if it produces deceleration.

• A motor operates in two modes. Motoring and braking.

• In motoring, it converts electrical energy to mechanical energy, which

supports its motion.
• In braking, it works as a generator; opposing motion.

• A motor can provide motoring and braking operations for both

forward and reverse directions.
• The inertia or dynamic torque appears when the speed changes
from one value to another. If the drive is undergoing
acceleration, this torque opposes drive motion.

• If the drive is being braked, supports motion. The inertia torque

both in magnitude and in sign, is determined as the algebraic
difference between the motor torque and the load torque. In
general the torque equation is written as;

d
 T  TL  J
dt
Selection of the sign to be placed before each of the torque depends on the
operating conditions and on the nature of Load torque.
 Power developed by a motor is given by the
product of speed and torque.

In quadrant I, power Developed is positive;

machine works as a motor supplying
Mechanical energy.

In quadrant II, power is negative. Hence the

machine works under Braking opposing the
motion.

Multi-quadrant operation of drives

 Referring Load Torques and Moment of Inertia
• Different parts of a load may be coupled through different
mechanisms, such as gears, V-belts, and crankshafts.
• These parte may have different speeds and different types of
motions such as rotational and translational.

Transmission links provided between a motor the machine it drives.

 Referring forces and translating masses to
a rotating shaft
• In many machines some of the moving parts rotate while others
go through a translational motion. A motion of translation such
as hoists, cranes, etc will have to be referred to a rotating shaft.

Gearing diagram of a hoist drive

 ii. Control of Electrical Drives and selection.

1. Modes of operation
An electrical drive operates in three modes.
a) Steady State
b) Acceleration including starting
c) Deceleration including stopping.

•steady state operation takes place when the

motor torque equals the load torque. i.e.

d m
T  Tl  J
dt
• The steady state operation for a given speed is realized by the
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.
ωm Tl
When the motor parameters are
adjusted to provide speed torque curve 1,
ωm1
1 Drive runs at the desired speed ωm1.

ωm2 Speed is changed to ωm2 when the

2 motor parameters are Adjusted to
provide speed- torque curve 2.

0
T
Principle speed control
• Acceleration and deceleration modes are transient operations.

• 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.

• Increase in motor torque is accompanied by an increase in

motor current. Care must be taken to restrict the motor current
within 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.
• Motor operation in deceleration mode is required when a
decrease in its speed is required. Deceleration occurs when load
torque exceeds the motor torque.

• In those applications where load torque is always present with

substantial magnitude, enough deceleration can be achieved by
simply reducing the motor torque to zero.

• In those applications where load torque may not always have

substantial amount or where simply reducing the motor torque
to zero does not provide enough deceleration, mechanical or
electrical brakes may be employed to produce the required
magnitude of deceleration.
• During electric braking, motor current tends to exceed the safe
limit. Appropriate changes are made to ensure that the current
is restricted within safe limits.

• Usually, the electric braking is employed in applications

requiring frequent, quick, accurate or rapid emergency stops.
eg. Suburban electric trains, etc
 2. SPEED CONTROL AND DRIVE CLASSIFICATIONS

Constant speed or single speed drives: - are drives where the driving
motor runs at a nearly fixed speed.

Variable speed drives: - are drives needing step less change in speed and
multispeed drives.

Multi-motor drive: - is when a number of motors are fed from a common

converter, or when a load is driven by more than one motor.

• Speed range a variable speed drive depends on the application.

In some applications, it can be from rated speed to 10% of rated
speed. In some other applications, speed control above rated speed is
also desired and the ratio of maximum to minimum speed can be as
high as 200. there are also applications where the range is as low as
from rated speed to 80% of rated speed.
• A variable speed drive is called constant torque drive if the
d i e’s a i u to ue apa ilit does ot ha ge ith
change in speed setting. The corresponding mode ( or region)
of operation is called constant torque mode.
• The constant power drive and constant power mode (or region)
are defined in the same way.
• Ideally, it is desired that for a given speed setting, the motor
speed should remain constant as load torque is changed from
no load to full load.
• In practice, speed drops with an increase in the load torque.
Quality of speed control system is measured in terms of speed
regulation.
3. Open loop control

In open loop control, the control action from

the controller is independent of the "process
output" (or "controlled process variable").
A good example of this is a central heating
boiler controlled only by a timer, so that heat
is applied for a constant time, regardless of
the temperature of the building. The control
action is the switching on/off of the boiler.
The process output is the building
temperature.
 4. Closed Loop control Drives
• Feedback loops in electrical drive may be provided to satisfy one
or more of the following requirements:
i) Protection
ii) Enhancement of speed response
iii) To improve steady-state accuracy
• various closed loop configurations which usually find application
in electrical drive are the following. The converter and associated
control circuit is represented by a single box marked converter.

4.1. Current limit control: - This scheme is employed to limit the

converter and motor current below a safe limit during transient
operations. It has a current feedback loop with a threshold logic
circuit. As long as the current is within a set maximum value,
feedback does not affect operation of the drive.
Current limit control
• If current exceeds the set maximum value, feedback loop
becomes active and current is forced below the set maximum
value, which causes the feedback loop to become inactive again.

• The current fluctuates around a set maximum limit during the

transient operation until the drive condition is such that the
current does not have a tendency to cross the maximum value.

Eg. As is known, during starting of a motor, current will fluctuate

around the set maximum value. When close to steady state
operation point, current will not have tendency to cross the
maximum value, consequently, feedback loop has no effect on
the drive operation.
4.2. Closed loop Torque control: - The closed loop torque control
scheme finds application in battery operated vehicles, rail cars
and electric trains.
• Driver presses the accelerator to set torque reference T*.
Through closed loop control of torque, the actual motor torque
T follows torque reference T*.
• Speed feedback loop is present through the driver. By putting
appropriate pressure on the accelerator, driver adjusts the
speed depending on traffic, road condition, car condition and
speed limit.

. closed loop torque control

4.3. Closed loop speed control: - Closed loop speed control scheme is
widely used electrical drives. It employs an inner current control
loop within an outer speed loop. Inner current control loop is
provided to limit the converter and motor current or motor
torque below a safe limit.

Current input output

Characteristic for current
limiter

Closed loop speed control

• Inner current loop is also beneficial in reducing the effect of drive
performance of any non linearity present in convertor-motor
system.
The above Closed loop speed control operates as follows;
- An increase in reference speed ω*m produces a positive error Δωm.
-Speed error is processed through a speed controller and applied
to a current limiter which saturates even for a small speed error.
Consequently, limiter sets current reference for inner current
control loop at a value corresponding to the maximum allowable
current. Drive accelerates at the maximum allowable current.
When close to the desired speed, limiter desaturates. Steady
state is reached at the desired speed (with some steady-state
error) and at current for which motor torque is equal to load
torque.
• A decrease in reference speed ω*m produces a negative speed
error. Current limiter saturates and sets current reference for inner
current loop at a value corresponding to the maximum allowable
current. Consequently, drive decelerates in braking mode at the
maximum allowable current.

•When close to the required speed, current limiter desaturates. The

operation is transferred from braking to motoring. Drive then
settles at a desired speed at current for which motor torque equals
the load torque.
4.7. Closed loop position control
• A closed loop position control consists of a closed loop speed
control system with an inner current control loop inside an
outermost position loop. Current and speed loop restrict the
current and speed within safe limits, enhance the speed of
response, reduce the effects of nonlinearities in the converter,
motor and load on the transient and steady state performance of
the position control system.
e.g. feed drive in machine tools, screw down mechanism in
rolling mills, etc.

Closed loop position control

 iii) Selection of motor power rating

• The power rating of a motor for a specific application must be

carefully chosen to achieve economy with reliability.
- use of a motor having insufficient rating
- it fails to drive the load;
- lowers the productivity and reliability through
frequent damages and shut-downs due to
overloading of the motor and power modulator;
- use of motor having extra power than needed
- extra initial cost
uneconomical
- extra loss of energy
Especially, IMs &
- operate below rated power factor Synch. motors
• It is clear that, heat is produced due to losses in copper, iron and
friction during motor operation.

• As the temperature of the motor increases beyond ambient value

part of the heat flows to the surrounding medium.

• As outflow of heat increases, equilibrium sets when the heat

generated becomes equal to heat dissipated.

• The motor temperature reaches steady state and it depends on

power loss, which intern depends on the output power of the
machine; and this dependence of temperature rise on output
power which is taken as thermal loading on the machine.
• Steady state temperature is not the same at various parts of the
machine. It is usually highest in the winding because loss
density in conductors is high and dissipation rate is low. (since
windings are embedded in slots)
• Among the various materials used in machines, insulation has
the lowest temperature limit.
• Depending on the temperature limits, insulating materials are
divided into classes.
Insulation class temperature limit
90oC
A 105oC
E 120oC
B 130oC
Insulation temperature limits
F 155oC
H 180oC
C Above 180oC
• For a specific operation, motor rating should be chosen such that
the insulation temperature never exceeds the prescribed limit.
Otherwise, it will lead to its immediate thermal break down
causing short circuit and damage of the winding.

• For loads which operate at a constant power and speed,

determination of motor power rating is simple.

• Usually, most loads operate at variable power and speed, and

the patterns of these variations are different for different
applications
Thermal model of motor for heating and cooling
• An accurate prediction of heat flow and temperature rise inside
an electrical motor is very difficult owing to complex geometrical
shapes and use of heterogeneous materials.

• A d i e e gi ee ’s dut is to select a motor rating for a given

application ensuring that temperature in various parts of motor
body do not exceed the safe limit.

• Thus, assuming that heat conductivities of various materials do

not differ by large amount and thus by assuming the motor to
be homogeneous, a simple thermal model of the machine can be
obtained. Although inaccurate, such a model is good enough to
determine rating of a motor.
• In self cooled motors where cooling fan is mounted on the
motor shaft, the velocity of cooling air varies with motor speed,
thus varying cooling time constant τ’.
• In high performance , medium and high power variable speed
drives, motor is always provided with separate forced cooling, so
that motor cooling be independent of speed.

This figure shows the variation

Of motor temperature rise with time
during heating and cooling. Thermal
time constants of a motor are far larger
than electrical and mechanical time
constants.

Heating and cooling curves

Classes of motor duty
• The three basic classes of motor duty are: -

a) Continuous duty

It denotes the motor operation at

a constant load torque for s duration
long enough for the motor temperature
To reach steady state value. This duty is
characterized by a constant motor loss.

Paper mill drives, compressors, fans etc

motor load diagram

b) Short time duty

Time of drive operation is considerably

less than the heating time constant and
motor is allowed to cool off to ambient
temperature before the motor is
required to operate again.
In this operation, the machine can be
overloaded until temperature at the
end of loading time reaches the
permissible limit.

e.g. Cranes, valve drives, drives for

position control, etc
Motor load diagram
c) Intermittent periodic duty
It consists of periodic duty cycles,
each consisting of a period of running
At a constant load and a rest period.
Neither the duration of running period
is sufficient to rise the temperature to a
Steady value nor the rest period is long
enough for the machine to cool to ambient
temperature.
e.g. metal cutting and drilling tool drives,
fork lift drives etc
motor load diagram
Determination of Motor rating
• To determine motor rating, duty cycles discussed previously can
be broadly classified as: -
a) Continuous duty
b)Fluctuating loads
c)Short-time and intermittent duty

a) Continuous duty: -
• The selection of the motor capacity for such a drive is simple if
we know approximately the constant power input required by
the machine.
• In order to select a motor for a rating equal to the known
power input requirement , one may be sure that this rating is
the maximum permissible with respect to heating because the
manufacturer designs and rates it so as to attain maximum
utilization of the material used at rated output.
• For continuous duty at constant or slightly varying load, a motor
is selected from the relevant catalogue for a capacity
corresponding to the power required.
• Although losses during starting are greater than those under rated
load, they may be neglected because starting under continuous
duty is very infrequent and practically no influence on motor
heating.
b) Fluctuating and intermittent duty
- Equivalent current, torque and power methods

- This method is based on approximation that the actual variable

motor current can be replaced by an equivalent Ieq, which
produces same losses in the motor as actual current. The
equivalent current is determined as follows;
- As we know, motor loss (Pl) consists of two components.
- Pc - Constant loss which is independent of load
- Core loss and friction loss
- Pcu - Variable loss () dependant on load
- copper loss
Thus for a fluctuating load consisting of n values of motor
currents I1, I2, I3, ….., In for duration t1, t2, t3, …., tn respectively,

     
the equivalent current Ieq is: -

P I     
Pc  I R
2 2 2
R t P I R t ... P I n R tn

t1  t 2  ...  t n
2 c eq 1 c eq 2 c
eq

Pc  I eq R 
P t  t  ...  t  
I 2
t 
 I 2
t  ...  I 2
ntn R 
t1  t 2  ...  t n t1  t 2  ...  t n
2 c 1 2 n 11 2 2
 I t  I 2t 2  ...  I n t n
I eq 
2 2 2

t1  t 2  ...  t n
11

I I2
I4

I1 In
I3

0
t
t1 t2 t3 t4 tn

Load diagram of Fluctuating load

 - If the current varies smoothly over a period T, Ieq is: -

I eq  
T
1 2
i dt
T 0
I
Load diagram
of Fluctuating load

0
t
- After Ieq is determined, a motor with next higher current rating
from commercially available rating is selected. Then this rating
is checked for its practical feasibility .
DC motor
- This motor can be allowed to carry larger than the rated current
for a short duration. ( Short time overload capacity of the motor)
- A normally designed DC machine is allowed to carry up to 2
times the rated current.
Let the ratio of maximum allowable current (short time overload
current capacity) to rated current be λ.


I max Where, Imax is the maximum value of current
and Irated is the motor rated current
I rated
If the above condition is not satisfied, the motor rating is calculated from: -

I rated 

I max
Induction and Synchronous motors
- In case of induction and synchronous motors, for stable
operation, maximum load torque should be well within the
breakdown torque of the motor.

- The ratio of breakdown to rated torque for induction motors

with normal design varies from 1.65 to 3 and for synchronous
motors it varies from 2 to 2.25
If the ratio of breakdown to rated torque is λ’, the the oto
torque rating is chosen based on;



Tmax
Trated
- The e ui ale t u e t ethod assu es o sta t losses to
remain constant for all operating points. Therefore, this method
should be carefully employed when these losses vary.
- It is also not applicable to motors with frequency (or speed)
dependent parameters of the equivalent circuit;
e.g. in deep bar and double squirrel cage rotor motors where
the rotor winding resistance and reactance vary widely
during starting and braking.
-When torque is directly proportional to current;
e.g. dc separately excited motor,

T t  T 2t 2  ...  T n t n
Teq 
2 2 2

t1  t 2  ...  t n
11
- When a motor operates at nearly fixed speed, its power will be
directly proportional to torque. Hence, for nearly constant
speed operation, power rating of the motor can be obtained
directly from;

p 1 t1  p 2t 2  ...  p n t n
Peq 
2 2 2

t1  t 2  ...  t n

Example
A constant speed drive has the following duty cycle.
i) Load rising from 0 to 400 KW ; 5 min
ii) Uniform load of 500 KW; 5 min
iii) Regenerative power of 400 KW returned to the supply; 4 min
iv) Remains idle for ; 2 min

Estimate power rating of the motor. Assume losses to be proportional to (power)2

Sol.
Rated power = rms value power. = Prms ;
Now the rms value of the power in interval (i) is;


1  400 
Pi   x  dx 
5 2
400
5 0 5 
KW
3

 400 
   5  5002  5  4002  4
2

Prms   3
 367KW
16

Since Pmax = 500 KW is less than two times Prms, motor rating = 367 KW
c) Short time duty
- In short time duty, time of motor operation is considerably less
than the heating time constant and motor is allowed to cool
down to the ambient temperature before it is required to
operate again.
- A motor with continuous duty power rating Pr can be overloaded
by a factor K ( K>1) such that the power rating becomes KPr and the
maximum temperature rise reaches the permissible value (θper).

θper

a
θ vs t curves for short time duty loads
(a) - with power KPr
(b) - with power pr b
0
t
 Intermittent periodic duty
- During a period of operation, if the speed changes in wide limits,
leading to changes in heating and cooling conditions, methods of
equivalent current, torque or power described previously can
not be employed.
- Let us consider an intermittent load where the motor is
alternatively subjected a fixed load Pr’ of du atio tr and stand
still condition of ts.
- As the motor is subjected to a periodic load, after the thermal
steady state is reached the temperature rise will fluctuate
between a max. value θmax and a minimum value θmin.
- For this load, the motor rating should be selected such that,
θmax ≤ θper where θper is the max. permissible temperature rise
of the motor.
- At the end of working (running) max. temperature will be;
tr tr

 max   ss (1  e  )   mine 
r r ……………….. .

And fall in temperature rise at the end of standstill interval ts will be

ts

 min   max e s …………………………… .

T
P’r P’r tr and ts are thermal time constants
of motor for working and standstill
θ
θmax θ

θmin Fig.5.7. intermittent periodic load

tr ts t
  ss 1  e t r /  r  t s / s 
Combining 5.22 and 5.23, we can get;


 max 1 e t / r
…………………….. .
r

For full utilization of the motor, θmax = θper . Further θper will be the motor

 ss
temperature rise when it is subjected to its continuous rated power Pr .


 max P1r
P1s ……………………….. .

Overload factor K is given by

K    1
1  e tr / r  t s / s 

1 e t r /  r
………… .
Example
5.2. A motor has a heating time constant of 60 min and cooling
time constant of 90 min. When run continuously on full load
of 20 KW, the final temperature rise is 40oC.
i) What load the motor can deliver for 10 min if this is followed
by a shut down period long enough for it to cool.
ii) If it is on an intermittent load of 10 min followed by 10 min
shut down, what is the maximum value of load it can supply
during the on load period.

Sol.
Alpha(α) is assumed to be zero, since constant and copper losses
are not available separately.
 i) When α = 0, the overloading factor is;

K   2.25
1 1
1 e t r / 
1 e 10 / 60

Permitted load = 2.25x20 = 51 KW

ii) From equation 5.26 for α = 0,

 10 10 
  
1  e tr / r  t s / s  1 e
 60 90 
K    1.257
0.2425
1 e t r /  r
1 e 10 / 60
0.1535
Permitted load = 1.257x20 = 25.14 KW
Work out
2. Half hour rating of a motor is 100 KW. Heating time constant is
80 min and the maximum efficiency occurs at 70% full load.
Determine the continuous rating of the motor.

3. A motor operates on a periodic duty cycle in which it is clutched

to its load for 10 min and declutched to run on no load for 20
min. Minimum temperature rise is 40oC. Heating and cooling
time constants are equal and have a value of 60 min. When
load is declutched continuously, the temperature rise is 15oC.
Determine,
i) Maximum temperature during the duty cycle.
ii) Temperature when the load is clutched continuously.
 6. DC motor drives
• Dc drives are widely used in applications requiring adjustable
speed, good speed regulations and frequent starting, braking and
reversing.
• Please revise the following topics in dc motor drives.
1. Commonly used dc motors and their performance
- separately excited, Shunt, Series, Compound,
Universal motor, Permanent magnet motor, etc

E  K em
- Basic equations applicable to all dc motors are:-

V  E  I a Ra motor
T  K eI a

 
K e K e 
V Ra
2
T
2. Starting
- Maximum current that a dc motor can safely carry during
starting is limited by the maximum current that can be
commutated with out sparking. Usually it is necessary to limit the
current to a safe value during starting.
3. Braking
- Regenerative braking
- Dynamic braking
- Plugging
4. Transient analysis different types of dc motors
5. Energy losses during transient operations
6. Speed control
- Armature voltage control
- field flux control
- Armature resistance control
Methods of armature voltage controls – Variable armature
voltage for speed control, starting, braking and reversing of
dc motors can be obtained by the following methods.

When the supply is AC,

- Ward- Leonard schemes
- Transformer with taps and uncontrolled rectifier bridge
- Static Ward- Leonard scheme or controlled rectifiers

When the supply is DC,

- Chopper control
 6.1. CHOPPER-CONTROLLED dc DRIVES
• Choppers also commonly known as dc-dc convertors, are used
to get variable dc voltage from a dc source of fixed voltage.

• A Chopper may be considered as dc equivalent of an ac

transformer since they behave in an identical manner. Like a
transformer, a chopper can be used to step down or step up the fixed
dc input voltage.

• Self commutated devices, such as MOSFETs, power transistors,

IGBTs, GTOs, and IGCTs, are usually used for building choppers.

• These devices are generally represented by a switch. When the

switch is off, no current can flow. Current flows through the
load when switch is o .
• The power semiconductor devices have on-state voltage drop of
0.5V to 2.5V across them. For the sake of simplicity, this voltage
drop across these devices is generally neglected.

PRINCIPLE OF CHOPPER OPERATION

• A hoppe is a high speed o " o off se i o du to s it h. It
connects source to load and disconnect the load from source at
a fast speed.

• In this manner, a chopped load voltage as shown in the following

Fig. is obtained from a constant dc supply of magnitude Vs.

• For the sake of highlighting the principle of chopper operation,

the circuitry used for controlling the on, off periods is not shown.
• During the period ton, chopper is on and load voltage is equal to
source voltage Vs. During the period toff, chopper is off, load
voltage is zero. In this manner, a chopped dc voltage is produced
at the load terminals

fig. 6.1. Chopper Circuit , Voltage and Current Waveform.

Average Voltage, Vo= (ton/ (ton+toff))*Vs = (ton/T)*Vs =αVs

ton = on-time.
toff = off-time.
T = ton + toff = Chopping period.
α = ton/toff.
Thus the oltage a e o t olled a i g dut le α.
Vo = f x ton x Vs
f = 1/T = chopping frequency.
CONTROL STRATEGIES
• The average value of output voltage Vo can be controlled
through duty cycle by opening and closing the semiconductor
switch periodically.
• The various control strategies for varying duty cycle are the
following:

1. Time ratio Control (TRC) and

2. Current-Limit Control.

Time ratio Control (TRC)

• In this control scheme, time ratio ton/T(duty ratio) is varied.
This is realized two different ways called Constant Frequency
System and Variable Frequency System as described below:
1. CONSTANT FREQUENCY SYSTEM
In this scheme, on-time is varied but chopping frequency f is kept
constant. Variation of ton means adjustment of pulse width, as such
this scheme is also called pulse-width-modulation scheme.

2. VARIABLE FREQUENCY SYSTEM

In this technique, the chopping frequency f is varied and either
(i) on-time ton is kept constant or
(ii) off-time toff is kept constant.
This method of controlling duty ratio is also called Frequency-
modulation scheme.
CURRENT- LIMIT CONTROL
• In this control strategy, the on and off of chopper circuit is
decided by the previous set value of load current. The two set
values are maximum load current and minimum load current.
• When the load current reaches the upper limit, chopper is
switched off and when the load current falls below lower limit,
the chopper is switched on.

• Switching frequency of chopper can be controlled by setting

maximum and minimum level of current. Current limit control
involves feedback loop, the trigger circuit for the chopper is
therefore more complex.

• PWM technique is the commonly chosen control strategy for

the power control in chopper circuit.
Chopper control of Separately excited Dc motor (Example)
Basics of Separately Excited DC Motor

As we can recall;
• Separately Excited DC motor has
field and armature winding with
separate supply.

• The field windings of the dc motor

are used to excite the field flux.

• Current in armature circuit is

supplied to the rotor via brush.

• The rotor torque is produced by

Fig. 6.2. separately excited dc motor interaction of field flux and armature
current.
• When a separately excited dc motor is excited by a field current
of if and an armature current of ia flows in the circuit, the motor
develops a back EMF and a torque to balance the load torque at
a particular speed.

• The field current if is independent of the armature current ia. Each

winding is supplied separately. Any change in the armature
current has no effect on the field current.

FIELD AND ARMATURE EQUATIONS

• Instantaneous field current

Vf  Rf if  Lf
di f Where Rf and Lf are field resistor and
inductor, respectively
dt
• Instantaneous armature current

Va  Ra ia  La  eg
dia
dt
Where, Ra and La are armature resistance and inductance respectively
eg = Kvωif - back emf (speed voltage)
Kv – constructional constant

BASIC TORQUE EQUATION

The Torque developed by the motor is

Tm  Kt i f ia  Ktia Where Kt = Kv - the torque constant

For normal operation, the developed torque must be equal to

d
the load torque plus the friction and inertia, i.e.

Tm  J  B  TL
dt
The STEADY-STATE TORQUE AND SPEED
The motor speed can be easily derived

Va  I a Ra

Kv I f
If Ra is a small value (which is usual), or when the motor is lightly
loaded, i.e. Ia is small;


Va That is, if the field current is kept
Constant, the motor speed depends
Kv I f only on the supply voltage
 Tm  Kt I f I a  B  TL
The developed torque is;

Pm  Tm
The required power is;

TORQUE AND SPEED CONTROL

From the above derivation important facts can be deduced for

steady-state operation of DC motor.

• Fo a fi ed field u e t, o flu If ) the torque demand can be

satisfied by varying the armature current (Ia).
• The motor speed can be controlled by:
– controlling Va (voltage control)
– controlling Vf (field control)
• These o se atio lead to the appli atio of a ia le DC oltage
for controlling the speed and torque of DC motor.
VARIABLE SPEED OPERATION

Fig 6.3. Torque Vs Speed Characteristic For Different Armature Voltages

Motoring control
• In a transistor chopper controlled separately excited dc motor,
transistor Tr is operated periodically with period T remains on for
a duration ton.
• wave forms of motor terminal voltage Va and armature current ia
for continuous conduction are shown below.

Fig. 6.4. Chopper control of separately excited motor

• Du i g o pe iod of the t a sisto , 0 ≤ t ≤ ton, the motor
terminal voltage is V. The operation is;

Ra ia  La  E V,
dia 0 ≤ t ≤ ton duty interval
dt
• In this interval, armature current increases from ia1 to ia2.
• At t = ton, Tr is turned off. Motor current freewheels through
diode DF and motor terminal voltage is zero during interval
ton ≤t ≤ T. f ee heeli g i te al . This i te al is des i ed the
equation;

Ra ia  La E0
dia
ton ≤ t ≤ T
dt
• Motor current decreases from ia2 to ia1 during this interval.
• Ratio of duty interval ton to chopper period T is called duty ratio
or duty cycle ( ).Thus,

 
Duty int erval ton
T T

From fig. 5.11. (b)

Va   Vdt  V
t
1 on
T 0

V  E
Ia 
Then

Ra
V
m   2T
Ra
K K
Regenerative braking
• Transistor Tr in a Chopper for regenerative braking is operated
periodically with a period T and on period of ton.
• The mechanical energy converted into electrical by the motor,
now working as a generator, partly increases the stored magnetic
energy in armature circuit inductance and remainder is
dissipated in armature resistance and transistor.

Fig. 6.5. Regenerative braking by chopper control ( Separ. Excited dc motor

• When Tr is turned off, armature current flows through diode D
and V, and reduces from ia2 to ia1.
• The i te al 0 ≤ t ≤ ton is now called energy storage interval and
interval ton ≤ t ≤ T the duty interval. Then the duty cycle is;

duty int erval T  ton

 
T T
From fig. 5.12.(b)

Va   Vdt  V
T
1
T ton

E  V
Then,

Ia 
Ra
 T   KI a
Since Ia is reversed,

V
  2T
Ra
K K

Dynamic braking

Fig.6.6. Circuit for dynamic braking of separately excited motor by chopper control
• Du i g the i te al 0 ≤ t ≤ ton, ia increases from ia1 to ia2. A part of
generated energy is stored in inductance and rest is dissipated in Ra
and Tr.
• During interval ton ≤ t ≤ T, ia decreases from ia2 to ia1. The energy
stored in inductance are dissipated in braking resistance RB, Ra and
diode D. Transistor Tr controls the magnitude of energy dissipated
in RB and therefore, controls its effective value.
• If ia is assumed to be rippleless dc, then energy consumed (EN) by

EN  I a RB (T  ton )
a chopper operation is;
2

• Average power consumed by RB is; P   I a RB (1   )

EN 2

T
 2  RB (1   )
P
•Effective value of RB is; RBE
Where,   on
t Ia
T
Example
1. A 230 V, 960 rpm and 200 A separately excited dc motor has an
armature resistance of 0.02 ohm. The motor is fed from a
chopper which provides both motoring and braking operations.
The source has a voltage of 230 V. Assuming continuous
condition,
i) Calculate duty ratio of the chopper for motoring operation at
rated torque and 350 rpm.
ii) Calculate duty ratio of the chopper for braking operation at
rated torque and 350 rpm.
iii) If max. duty ratio of the chopper is limited to 0.95 and max.
permissible motor current is twice the rated, calculate max.
permissible motor speed obtainable without field weakening
and power fed to the source.
iv) If motor field is also controlled in (iii), calculate field current as
a fraction of its rated value for a speed of 1200 rpm.
At rated operation; E  230  (200 0.02)  226V
Sol.

i)
E  226  82.4V
350
At 350 rpm;
960

Motor terminal voltage; Va  E  I a Ra  82.4  (200 0.02)  86.4V

Duty cycle;    0.376

86.4
230

ii) Va  E  I a Ra  82.4  (200 0.02)  78.4V

Duty cycle;    0.34

78.4
230
iii) Maximum available Va  0.95  230  218.5V
E  Va  I a Ra  218.5  (200 2  0.02)  226.5V

 960  962rpm
226.5
Max. permissible motor speed = 226

Assuming lossless chopper, power fed to the source;

P  Va I a  218.5  400  87.4 Kw

iv) As in (iii) E = 226.5 V for which at rated field current speed = 960
rpm. Assuming linear magnetic circuit, E will be inversely
proportional to field current. Field current as a ratio of its rated
value = 960/1200 = 0.8
 6.3. Induction motor control.
What to control? Speed and Torque of the motor . These
parameters should be controlled to suit the load requirement.
• Availability of semiconductor devices (thyristors, power
transistors, IGBT, GTO, etc) allowed the development of variable
speed IM drives.
Let’s e all pe fo a e pa a ete s of IMs.
'
R1 X1 I '
2
X 2

I1 Io

'

V1 Xm
R
2

Pag

Fig. 6.7. Per phase rotor referred equivalent circuit of IM

 R1 X1 X’2
I1 Since stator impedance drop is
Io I’2 Generally negligible compared to
Terminal voltage, the equivalent
R2'
V1 Xm Circuit can be simplified.
S

Pag
Fig. 6.8. Simplified equiv. circuit.

• Slip is defined as

; N s 1  S   nr
N s  nr
S ---------- 6.3.1.

ms  m ω ms, ωm are synchronous and rotor angular speeds

Ns

S
ms
OR

4f
Where, ms 

m  ms (1  S )
p
------------------------------ 6.3.2
1. I ' 
  j  X 1  X 2 '
V
 R2 ' 
 R1 
2 ------------------------------ 6.3.3

 s 

2. Pg  3I '2 2
R2 ' ----------------------------------- 6.3.4.
s
3. Pcu  3I 2 R2
'2 '
----------------------------------- 6.3.5.

1 s 
4. Pm  Pg  Pcu  3I 2 R2  
 s 
'2 '
--------------------- 6.3.6.

T 
m ms
Pm 3 '2 R2 '
5. I2 ------------------------ 6.3.7
s
 
 2 R2 '

3  
T
V
m   
   X 1  X 2 ' 
s
R2 ' 
  R1 
2
----------- 6.3.8

 s  
2
Speed control of IM
• The following methods are employed for speed control of IM.
i) Pole changing
ii) Stator voltage control
iii) Supply frequency control
iv) Eddy current coupling
v) Rotor resistance control
vi) Slip power recovery
we will focus on the most commonly used ones. (ii, iii, v)

6.3.1. Stator voltage control

• By reducing stator voltage, speed of IM can be reduced by an
amount which is sufficient for the speed control some drives.
• we recall that, while developed torque is proportional to voltage
square, current is proportional to voltage.
• As voltage is reduced to reduce speed, for the same current
motor develops lower torque. Thus, this method is suitable for
applications where torque demand reduces with speed. (fan, pump
drives)
ωm

ωms TL

V
V’
0
T
fig. 6.9. Stator voltage control
• If stator copper loss, core loss, and friction windage loss are
ignored, then motor efficiency is;

  (1  s )
Pm ------------------- 6.3.9
Pg
• The above equation shows that, the efficiency falls with
decrease in speed. The speed control is essentially obtained by
dissipating a portion of rotor input power in rotor resistance.
Thus, not only the efficiency is low, the power dissipation occurs
in rotor itself, which may overheat the rotor .
• Because of these reasons, this drive is employed in fan and
pump drives of low power ratings and for narrow speed range.

• Variable voltage for speed control is obtained using ac voltage

controllers.
Control by ac voltage controllers
• Domestic fan motors, which are always single-phase , are
controlled by a single phase triac voltage controller.
- Speed control is obtained by varying firing angle of the triac.
- These solid state regulators are preferred over conventional
variable resistance regulators because of higher efficiency.

fig. 6.10. Stator voltage control by semiconductor voltage controller

• In industries, commonly 3-phase motors are used and for their
speed control, thyristor voltage controllers are employed.
- speed control is obtained by varying conduction period of
thyristors.
- The power factor of an ac regulator is;

PF  cos1
I1
---------------------- 6.3.10
I rms

Where, I1 – fundamental component of source current, A Distortion factors

Irms – rms source current, A
φ1 – phase difference b/n Voltage and current ; (displacement factor)

•With increase in firing angle, both distortion factor and

displacement factor reduce giving a low PF.
Workout

1. A 2.8 KW, 400 V, 50 hz, 4 pole, 1370 rpm, delta connected

squirrel cage IM has the following parameters referred to the
stator: Rs = 2Ω, Rr’ = Ω, Xs = Xr’ = Ω, Xm = 80Ω. Motor speed is
controlled by stator voltage control. When driving a fan load, it
runs at rated speed at rated voltage. Calculate
i) motor terminal voltage, current and torque at 1200 rpm
ii) motor speed, current and torque for the terminal voltage of
300 V.

Sol.